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 = 500007;
  stack<long long> s;
  long long a[N];
  long long f[N];
  long long sum[N];
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    s.push(0);
    f[0] = 1;
    sum[0] = 1;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      while (!s.empty() && a[s.top()] >= a[i]) {
        long long pos = s.top();
        s.pop();
        if (s.empty()) {
          ans = (ans - a[pos] * sum[pos - 1] % mod + mod) % mod;
        } else {
          ans = (ans - a[pos] * (sum[pos - 1] - sum[s.top() - 1] + mod) % mod +
                 mod) %
                mod;
        }
      }
      if (s.empty()) {
        ans = (ans + a[i] * sum[i - 1] % mod) % mod;
      } else {
        ans = (ans + a[i] * ((sum[i - 1] - sum[s.top() - 1] + mod) % mod) % mod) %
              mod;
      }
      s.push(i);
      if (i % 2 == 1) {
        f[i] = ans;
      } else {
        f[i] = (mod - ans) % mod;
      }
      if (i % 2 == 1) {
        f[i] = (mod - f[i]) % mod;
      }
      sum[i] = (f[i] + sum[i - 1]) % mod;
    }
    if (n % 2 == 1) {
      printf("%lld\n", (mod - f[n]) % mod);
    } else {
      printf("%lld\n", f[n]);
    }
    return 0;
  }
  