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;
  int n, top;
  long long x, ans;
  const int N = 2e5 + 5;
  const int mo = 998244353;
  long long a[N], f[N], sum[N], g[N], h[N];
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    f[0] = 1;
    sum[0] = 1;
    for (int i = 1; i <= n; i++) {
      while (a[i] <= a[g[top]]) top--;
      if (top)
        x = sum[i - 1] - sum[g[top] - 1];
      else
        x = sum[i - 1];
      x = -x * a[i] % mo;
      f[i] = (h[top] + x) % mo;
      sum[i] = (sum[i - 1] + f[i]) % mo;
      g[++top] = i;
      h[top] = (h[top - 1] + x) % mo;
    }
    ans = f[n];
    if (n % 2) ans = -f[n];
    ans = (ans % mo + mo) % mo;
    printf("%d\n", ans);
  }
  