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>
  int main() {
    static int aa[200000], dp[200000 + 1], dq[200000], dr[200000], qu[200000];
    int n, i, cnt;
    scanf("%d", &n);
    for (i = 0; i < n; i++) scanf("%d", &aa[i]);
    dp[0] = n % 2 == 0 ? 1 : 998244353 - 1;
    cnt = 0;
    for (i = 0; i < n; i++) {
      int q = dp[i];
      while (cnt && aa[qu[cnt - 1]] >= aa[i]) {
        q = (q + dq[cnt - 1]) % 998244353;
        cnt--;
      }
      dq[cnt] = q, qu[cnt] = i,
      dr[cnt] = ((cnt == 0 ? 0 : dr[cnt - 1]) + (long long)q * aa[i]) % 998244353,
      cnt++;
      dp[i + 1] = (998244353 - dr[cnt - 1]) % 998244353;
    }
    printf("%d\n", dp[n]);
    return 0;
  }
  