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 INF = 0x3f3f3f3f;
  const long long mod = 998244353;
  const int N = 1000005;
  int n, a[N];
  long long all, f[N], sum[N];
  int tp, s[N];
  void update(int x) {
    while (tp && a[x] <= a[s[tp]]) {
      all = (all - a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
            mod;
      --tp;
    }
    s[++tp] = x;
    all = (all + a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
          mod;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 2; i <= n + 1; i++) scanf("%d", &a[i]);
    f[1] = sum[1] = 1;
    for (int i = 2; i <= n + 1; i++) {
      update(i);
      f[i] = (((i % 2 == 0) ? all : -all) + mod) % mod;
      sum[i] = (sum[i - 1] + (i % 2 ? 1 : -1) * f[i] % mod + mod) % mod;
    }
    long long ans = f[n + 1];
    ans = (ans % mod + mod) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  