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 int mod = 998244353;
  int mult(int a, int b) { return (1LL * a * b) % mod; }
  int sum(int a, int b) {
    int s = a + b;
    if (s >= mod) s -= mod;
    return s;
  }
  int sub(int a, int b) {
    int s = a - b;
    if (s < 0) s += mod;
    return s;
  }
  int n;
  const int maxN = 5e5 + 10;
  int a[maxN];
  int dp[maxN];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    dp[0] = 1;
    vector<pair<int, int>> st;
    int cur = 0;
    for (int i = 1; i <= n; i++) {
      int to_add = 0;
      while (!st.empty() && a[st.back().first] >= a[i]) {
        int was_sum = mult(st.back().second, a[st.back().first] % mod);
        cur = sub(cur, was_sum);
        to_add = sum(to_add, st.back().second);
        st.pop_back();
      }
      int p = dp[i - 1];
      if ((i - 1) % 2 == 1) p = sub(0, p);
      to_add = sum(to_add, p);
      cur = sum(cur, mult(to_add, a[i] % mod));
      st.emplace_back(i, to_add);
      if (i % 2 == 1) {
        dp[i] = cur;
      } else {
        dp[i] = sub(0, cur);
      }
    }
    cout << dp[n];
    return 0;
  }
  