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 dim = 500009, mod = 998244353;
  struct elem {
    long long x, s;
  } stiva[dim];
  long long vf = 1;
  long long n, a[dim];
  long long dp[dim], sum[dim];
  signed main() {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    dp[0] = 1;
    sum[1] = a[1] % mod;
    stiva[vf] = {a[vf], vf};
    for (long long i = 1; i <= n; i++) {
      long long x = a[i + 1], s;
      if (i % 2 == 1) {
        dp[i] = sum[vf];
        s = (mod - dp[i]) % mod;
      } else {
        dp[i] = (mod - sum[vf]) % mod;
        s = dp[i];
      }
      while (vf && stiva[vf].x > x) {
        s += stiva[vf--].s;
        s %= mod;
      }
      stiva[++vf] = {x, s};
      sum[vf] = sum[vf - 1] + x * s;
      sum[vf] %= mod;
    }
    cout << dp[n];
  }
  