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 MAXN = 5e5 + 10;
  const int P = 998244353;
  int N, A[MAXN], sum[MAXN], tot, f[MAXN];
  int stk[MAXN], tp;
  inline void add(int& a, const int& b) { a = a + b >= P ? a + b - P : a + b; }
  inline void sub(int& a, const int& b) { a = a - b < 0 ? a - b + P : a - b; }
  inline int psum(const int& a, const int& b) {
    return a + b >= P ? a + b - P : a + b;
  }
  inline int pdif(const int& a, const int& b) {
    return a - b < 0 ? a - b + P : a - b;
  }
  int main() {
    scanf("%d", &N);
    for (int i = (1), iend = (N + 1); i < iend; ++i) scanf("%d", &A[i]);
    f[0] = N & 1 ? P - 1 : 1;
    sum[0] = f[0];
    for (int i = (1), iend = (N + 1); i < iend; ++i) {
      while (tp && A[stk[tp - 1]] > A[i]) {
        int L = tp >= 2 ? stk[tp - 2] : 0, R = stk[tp - 1] - 1;
        sub(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[stk[tp - 1]] % P);
        tp--;
      }
      int L = tp ? stk[tp - 1] : 0, R = i - 1;
      add(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[i] % P), stk[tp++] = i;
      f[i] = (tot ? P - tot : tot), sum[i] = sum[i - 1], add(sum[i], f[i]);
    }
    printf("%d\n", f[N]);
    return 0;
  }
  