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 mod = 998244353, N = 500005;
  long long a, b, c, d, e, f, g = 1, h[N], arr[N], dp[N][2], pre[N][2];
  string s;
  vector<long long> v;
  void solve() {
    cin >> a;
    for (long long i = 1; i <= a; i++) {
      cin >> arr[i];
    }
    stack<long long> stc;
    dp[0][0] = pre[0][0] = 1;
    for (long long i = 1; i <= a; i++) {
      while (stc.size() and arr[stc.top()] >= arr[i]) stc.pop();
      dp[i][0] = (!stc.size()
                      ? (pre[i - 1][1] * arr[i]) % mod
                      : (dp[stc.top()][0] +
                         (pre[i - 1][1] - pre[stc.top() - 1][1] + mod) * arr[i]) %
                            mod);
      dp[i][1] = (!stc.size()
                      ? (pre[i - 1][0] * arr[i]) % mod
                      : (dp[stc.top()][1] +
                         (pre[i - 1][0] - pre[stc.top() - 1][0] + mod) * arr[i]) %
                            mod);
      pre[i][0] = (dp[i][0] + pre[i - 1][0]) % mod;
      pre[i][1] = (dp[i][1] + pre[i - 1][1]) % mod;
      stc.push(i);
    }
    cout << (a % 2 ? (dp[a][1] - dp[a][0] + mod) % mod
                   : (dp[a][0] - dp[a][1] + mod) % mod)
         << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    while (g--) solve();
  }
  