description: You are given an array of n positive integers a_1, a_2, …, a_n. Your task is to calculate the number of arrays of n positive integers b_1, b_2, …, b_n such that: 
  
    * 1 ≤ b_i ≤ a_i for every i (1 ≤ i ≤ n), and 
    * b_i ≠ b_{i+1} for every i (1 ≤ i ≤ n - 1). 
  
  
  
  The number of such arrays can be very large, so print it modulo 998 244 353.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
  
  Output
  
  Print the answer modulo 998 244 353 in a single line.
  
  Examples
  
  Input
  
  
  3
  2 2 2
  
  
  Output
  
  
  2
  
  Input
  
  
  2
  2 3
  
  
  Output
  
  
  4
  
  Input
  
  
  3
  1 1 1
  
  
  Output
  
  
  0
  
  Note
  
  In the first test case possible arrays are [1, 2, 1] and [2, 1, 2].
  
  In the second test case possible arrays are [1, 2], [1, 3], [2, 1] and [2, 3].
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long mod = 998244353;
  int main() {
    int n;
    cin >> n;
    vector<long long> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<long long> dp(n + 2);
    vector<long long> sum(n + 2);
    vector<long long> ndp(n + 2);
    sum[0] = dp[0] = 0;
    sum[1] = dp[1] = 1;
    stack<int> st;
    st.push(0);
    a[0] = 0;
    for (int i = 1, idx; i <= n; i++) {
      idx = st.top();
      while (a[idx] >= a[i]) {
        st.pop();
        idx = st.top();
      }
      st.push(i);
      i++;
      idx++;
      ndp[i] = (ndp[idx] + (sum[i - 1] - sum[idx - 1]) * a[i - 1]) % mod;
      dp[i] = (i % 2 == 1 ? -1 : 1) * ndp[i];
      sum[i] = (sum[i - 1] + (i % 2 == 1 ? 1 : -1) * dp[i]) % mod;
      i--;
    }
    cout << (dp[n + 1] + mod) % mod << '\n';
    return 0;
  }
  