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 MX = 200005, mod = 998244353;
  int n;
  long long a[MX], dp[MX];
  stack<vector<long long> > st;
  void main_() {
    cin >> n;
    for (int i = 0; i < int(n); i++) cin >> a[i];
    long long acu = 0;
    dp[n] = 1;
    for (int i = n - 1; i >= 0; i--) {
      long long s = i % 2 ? -dp[i + 1] : dp[i + 1];
      while (st.size() && st.top()[0] >= a[i]) {
        (acu -= st.top()[0] * st.top()[1]) %= mod;
        (s += st.top()[1]) %= mod;
        st.pop();
      }
      (acu += a[i] * s) %= mod;
      st.push({a[i], s});
      dp[i] = i % 2 ? -acu : acu;
    }
    cout << ((((dp[0]) % (mod)) + (mod)) % (mod)) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) main_();
    return 0;
  }
  