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 P = 998244353;
  int N;
  long long A[200009];
  long long dp[200009][2];
  long long s[200009][2];
  long long res;
  void In() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
  }
  inline long long Mod(long long x) { return ((x % P) + P) % P; }
  void Solve() {
    int stk[200009];
    int tt = 0;
    int t;
    dp[0][0] = s[0][0] = 1;
    for (int i = 1; i <= N; i++) {
      while (tt && A[stk[tt]] >= A[i]) tt--;
      t = stk[tt];
      if (!tt)
        dp[i][0] = Mod(s[i - 1][1] * A[i]), dp[i][1] = Mod(s[i - 1][0] * A[i]);
      else
        dp[i][0] = Mod(dp[t][0] + Mod(A[i] * Mod(s[i - 1][1] - s[t - 1][1]))),
        dp[i][1] = Mod(dp[t][1] + Mod(A[i] * Mod(s[i - 1][0] - s[t - 1][0])));
      s[i][0] = Mod(s[i - 1][0] + dp[i][0]),
      s[i][1] = Mod(s[i - 1][1] + dp[i][1]);
      stk[++tt] = i;
    }
  }
  void Out() {
    printf("%lld\n", Mod((dp[N][0] - dp[N][1]) * ((N & 1) ? -1ll : 1ll)));
  }
  int main() {
    In();
    Solve();
    Out();
    return 0;
  }
  