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 = 200005;
  const int mod = 998244353;
  int cursum, a[maxN], dp[maxN], sum[maxN];
  void add(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  inline int get(int l, int r) {
    int ret = sum[r];
    if (l == 0) return ret;
    if ((r - l) & 1) {
      add(ret, mod - sum[l - 1]);
    } else {
      add(ret, sum[l - 1]);
    }
    return ret;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    stack<int> inds;
    a[0] = 0;
    dp[0] = 1;
    sum[0] = 1;
    inds.push(0);
    for (int i = 1; i <= n; i++) {
      while (a[inds.top()] > a[i]) {
        inds.pop();
      }
      int j = inds.top();
      dp[i] = get(j, i - 1) * 1LL * a[i] % mod;
      if (j > 0) {
        if (i % 2 == j % 2)
          add(dp[i], dp[j]);
        else
          add(dp[i], mod - dp[j]);
      }
      sum[i] = dp[i] - sum[i - 1];
      if (sum[i] < 0) sum[i] += mod;
      inds.push(i);
    }
    cout << dp[n] << '\n';
    return 0;
  }
  