This problem requires an insightful observation into the relationship between interval boundaries and Cartesian tree topology.

Key Insight: Interval Transformation

First, transform the given intervals by incrementing each $l_i$ and decrementing each $r_i$. These adjusted ranges correspond precisely to the "influence intervals" in a Cartesian tree. Examining subtasks 3 and 4 reveals a crucial hint: they mention the existence of a valid permutation. If the problem merely required checking for existence, this partial score would be trivial. This suggests a stronger property: for any valid rearrangement of the $l_i$ and $r_i$ arrays, the Cartesian tree structure is uniquely determined.

Partial Solution: Tree DP on Fixed Structure

For subtasks 3 and 4, we assume the tree is built from the intervals. We need to assign numbers from the set $S = {1, 2, \ldots, n}$ to nodes such that the min-heap property holds. At any node $u$ with candidate set $S$, the smallest element must be placed at $u$, while the remaining elements are partitioned into subsets $S_1$ and $S_2$ for the left and right subtrees.

Let $dp_u$ represent the number of ways to fill a candidate set of size $size_u$ into the subtree rooted at $u$. The recurrence follows the multinomial coefficient:

$$dp_u = dp_{left_u} \times dp_{right_u} \times \binom{size_u - 1}{size_{left_u}}$$

The answer is $dp_{root}$. Below is an implementation using this approach with precomputed factorials and modular inverses:

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1000005;
const int MOD = 998244353;

long long fact[MAXN], invFact[MAXN], inv[MAXN];
vector<int> rightEndpoints[MAXN];
int n;

long long modComb(int a, int b) {
    if (b < 0 || b > a) return 0;
    return fact[a] * invFact[b] % MOD * invFact[a - b] % MOD;
}

long long calculate(int left, int right) {
    // Pop the current node's right boundary
    if (!rightEndpoints[left].empty()) 
        rightEndpoints[left].pop_back();
    
    if (left + 1 >= right - 1) return 1;
    
    int mid = rightEndpoints[left].empty() ? left + 1 : rightEndpoints[left].back();
    long long leftWays = calculate(left, mid);
    long long rightWays = calculate(mid, right);
    
    int totalSize = right - left - 2;
    int leftSize = mid - left - 1;
    
    return leftWays * rightWays % MOD * modComb(totalSize, leftSize) % MOD;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n;
    vector<int> L(n + 1), R(n + 1);
    for (int i = 1; i <= n; ++i) cin >> L[i];
    for (int i = 1; i <= n; ++i) cin >> R[i];
    
    // Precompute factorials and inverses
    fact[0] = invFact[0] = 1;
    inv[1] = 1;
    for (int i = 2; i <= n; ++i) 
        inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;
    for (int i = 1; i <= n; ++i) {
        fact[i] = fact[i - 1] * i % MOD;
        invFact[i] = invFact[i - 1] * inv[i] % MOD;
    }
    
    // Clear and populate interval structure
    for (int i = 0; i <= n; ++i) rightEndpoints[i].clear();
    for (int i = 1; i <= n; ++i) 
        rightEndpoints[L[i]].push_back(R[i]);
    for (int i = 0; i <= n; ++i) 
        sort(rightEndpoints[i].begin(), rightEndpoints[i].end());
    
    cout << calculate(0, n + 1) << endl;
    return 0;
}

This approach yields 55 points, confirming that the Cartesian tree structure is indeed fixed by the interval boundaries.

Full Solution: Reconstructing the Tree from Boundary Frequencies

Since the tree structure is unique for any valid rearrangement, we must reconstruct it using only the invariant properties: the frequency of each value appearing as a left or right boundary.

Consider the depth of each value in the Cartesian tree. When a value $a_i$ first appears as a right boundary, it indicates that $a_{i+1}$ becomes the minimum of a new interval, splitting the sequence at position $i$. The value $a_i$ continues to serve as a right boundary in ancestor intervals until it itself becomes a minimum (the root of its own subtree). Therefore, the difference in depths between adjacent elements $depth_i - depth_{i+1}$ exactly equals the number of times $i$ appears as a right boundary. Similarly, left boundary frequencies determine depth increases.

Since every number appears atleast once as a partition point, we can reconstruct the relative depths. Setting $depth_1 = 0$, we compute the depth array and then construct the Cartesian tree using a monotonic stack (maintaining a min-stack on depths). Finally, we apply the tree DP described earlier.

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1000005;
const int MOD = 998244353;

long long factorial[MAXN], invFactorial[MAXN], modularInv[MAXN];
int leftCount[MAXN], rightCount[MAXN];
int nodeDepth[MAXN], subtreeSize[MAXN];
int leftChild[MAXN], rightChild[MAXN];
int stackArray[MAXN], top;
int n;

void precompute() {
    factorial[0] = invFactorial[0] = 1;
    modularInv[1] = 1;
    for (int i = 2; i <= n; ++i)
        modularInv[i] = (MOD - MOD / i) * modularInv[MOD % i] % MOD;
    for (int i = 1; i <= n; ++i) {
        factorial[i] = factorial[i - 1] * i % MOD;
        invFactorial[i] = invFactorial[i - 1] * modularInv[i] % MOD;
    }
}

// Iterative post-order DP to avoid deep recursion
long long solveTree(int root) {
    if (root == 0) return 1;
    
    vector<pair<int, int>> postOrder;
    vector<int> stk;
    stk.push_back(root);
    
    while (!stk.empty()) {
        int u = stk.back();
        stk.pop_back();
        postOrder.push_back({u, 0}); // 0 indicates first visit
        
        if (rightChild[u]) stk.push_back(rightChild[u]);
        if (leftChild[u]) stk.push_back(leftChild[u]);
    }
    
    // Process in reverse for post-order
    reverse(postOrder.begin(), postOrder.end());
    vector<long long> dp(n + 2, 1);
    
    for (auto &[u, _] : postOrder) {
        subtreeSize[u] = 1;
        if (leftChild[u]) subtreeSize[u] += subtreeSize[leftChild[u]];
        if (rightChild[u]) subtreeSize[u] += subtreeSize[rightChild[u]];
        
        long long ways = dp[leftChild[u]] * dp[rightChild[u]] % MOD;
        int leftSize = leftChild[u] ? subtreeSize[leftChild[u]] : 0;
        int rightSize = rightChild[u] ? subtreeSize[rightChild[u]] : 0;
        
        // C(leftSize + rightSize, leftSize)
        long long comb = factorial[leftSize + rightSize] * invFactorial[leftSize] % MOD * invFactorial[rightSize] % MOD;
        dp[u] = ways * comb % MOD;
    }
    
    return dp[root];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int l; cin >> l;
        leftCount[l + 1]++;
    }
    for (int i = 1; i <= n; ++i) {
        int r; cin >> r;
        rightCount[r - 1]++;
    }
    
    precompute();
    
    // Reconstruct depths from boundary frequencies
    nodeDepth[1] = 0;
    for (int i = 2; i <= n; ++i) {
        if (leftCount[i]) nodeDepth[i] = nodeDepth[i - 1] + leftCount[i];
        else if (rightCount[i - 1]) nodeDepth[i] = nodeDepth[i - 1] - rightCount[i - 1];
        else nodeDepth[i] = nodeDepth[i - 1];
    }
    
    // Build Cartesian tree using monotonic stack on depths
    top = 0;
    for (int i = 1; i <= n; ++i) {
        int last = 0;
        while (top > 0 && nodeDepth[i] < nodeDepth[stackArray[top]]) {
            last = stackArray[top];
            top--;
        }
        if (top > 0) rightChild[stackArray[top]] = i;
        leftChild[i] = last;
        stackArray[++top] = i;
    }
    int root = stackArray[1];
    
    cout << solveTree(root) << endl;
    return 0;
}

This completes the solution. Note that there exists an alternative approach utilizing only the $l_i$ array, which would be an interesting extension to explore.