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 N = 5e5 + 10;
  const int mod = 998244353;
  int n;
  int a[N], dp[N];
  stack<pair<int, pair<int, int> > > s;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    dp[0] = 1;
    int res = 0;
    for (int i = 1; i <= n; i++) {
      int sum = 1ll * a[i] * dp[i - 1] % mod;
      int tmp = dp[i - 1];
      res = (res + sum) % mod;
      while (!s.empty() && s.top().first > a[i]) {
        sum = (sum + 1ll * s.top().second.first * a[i] % mod) % mod;
        res = (res - s.top().second.second + mod) % mod;
        res = (res + 1ll * s.top().second.first * a[i] % mod) % mod;
        tmp = (tmp + s.top().second.first) % mod;
        s.pop();
      }
      s.push(make_pair(a[i], make_pair(tmp, sum)));
      dp[i] = (mod - res) % mod;
    }
    int ans = dp[n];
    if (n & 1) ans = (mod - ans) % mod;
    cout << ans << endl;
    return 0;
  }
  