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 N = 2e5 + 10;
  const long long Mod = 998244353;
  int n;
  int a[N];
  long long dp[N], sum[N];
  struct date {
    int num[4];
  } nw;
  stack<date> st;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    T op = 1;
    char c = getchar();
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') op = -1;
    for (; c <= '9' && c >= '0'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= op;
  }
  int main() {
    read(n);
    for (int i = 1; i < n + 1; ++i) read(a[i]);
    dp[0] = 1;
    long long s = 0;
    st.push(nw);
    for (int i = 1; i < n + 1; ++i) {
      int l = i;
      long long tmp = 0;
      while (!st.empty() && st.top().num[0] > a[i]) {
        int L = st.top().num[1], R = st.top().num[2];
        s = (s - st.top().num[3]) % Mod;
        tmp = (tmp + a[i] * (sum[R] - sum[L - 1])) % Mod;
        l = L;
        st.pop();
      }
      tmp = (tmp + a[i] * dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
      nw.num[0] = a[i], nw.num[1] = l, nw.num[2] = i, nw.num[3] = (int)tmp;
      st.push(nw);
      dp[i] = (s + tmp) * (i & 1 ? -1 : 1) % Mod;
      sum[i] = (sum[i - 1] + dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
      s = (s + tmp) % Mod;
    }
    printf("%lld\n", (dp[n] + Mod) % Mod);
    return 0;
  }
  