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 Maxn = 2e5;
  const int Mod = 998244353;
  int a[Maxn + 5], l[Maxn + 5];
  int f[Maxn + 5], sum[Maxn + 5];
  int n;
  int getsum(int l, int r) {
    int val = (sum[r - 1] - sum[l - 1] + Mod) % Mod;
    if ((l & 1) ^ ((r - l + 1) & 1)) val = Mod - val;
    return val;
  }
  int main() {
    scanf("%d", &n);
    for (register int i = 1; i <= n; i++) scanf("%d", &a[i]);
    stack<int> s;
    for (register int i = 1; i <= n; i++) {
      while (!s.empty() && a[s.top()] > a[i]) s.pop();
      if (!s.empty()) l[i] = s.top();
      s.push(i);
    }
    f[0] = sum[0] = 1;
    for (register int i = 1; i <= n; i++) {
      int k = l[i];
      if (k) f[i] = (i - k) & 1 ? Mod - f[k] : f[k];
      f[i] = (f[i] + 1ll * a[i] * getsum(k, i) % Mod) % Mod;
      sum[i] = (sum[i - 1] + (i & 1 ? Mod - f[i] : f[i])) % Mod;
    }
    printf("%d\n", f[n]);
    return 0;
  }
  