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;
  const int N = 500000;
  int a[N + 5];
  long long sum[N + 5], f[N + 5];
  int main() {
    stack<array<int, 4>> s;
    s.push({0, 0, 0, 0});
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long tot = 0;
    f[0] = 1;
    for (int i = 1; i <= n; i++) {
      int left = i;
      long long temp = 0;
      while (!s.empty() && s.top()[0] >= a[i]) {
        auto now = s.top();
        s.pop();
        tot = (tot - now[3] + mod) % mod;
        temp = (temp +
                (long long)a[i] * (sum[now[2]] - sum[now[1] - 1] + mod) % mod) %
               mod;
        left = now[1];
      }
      long long coef = i % 2 ? mod - 1 : 1;
      temp = (temp + (long long)a[i] * coef % mod * f[i - 1] % mod) % mod;
      s.push({a[i], left, i, (int)temp});
      tot = (tot + temp) % mod;
      f[i] = tot * coef % mod;
      sum[i] = (sum[i - 1] + f[i - 1] * coef % mod) % mod;
    }
    printf("%lld", f[n]);
    return 0;
  }
  