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 + 2, p = 998244353;
  int a[N], g[N], st[N], ys[N];
  int n, tp, nowtot;
  void inc(int &x, int y) {
    if ((x += y) >= p) x -= p;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int i, j;
    cin >> n;
    for (i = 1; i <= n; i++) cin >> a[i];
    g[0] = 1;
    for (i = 1; i <= n; i++) {
      int now = 0;
      while (tp && a[st[tp]] >= a[i]) {
        inc(now, ys[tp]);
        nowtot = (nowtot + (long long)(p - a[st[tp]] + a[i]) * ys[tp]) % p;
        --tp;
      }
      st[++tp] = i;
      nowtot = (nowtot + (long long)g[i - 1] * a[i]) % p;
      g[i] = (p - nowtot) % p;
      inc(now, g[i - 1]);
      ys[tp] = now;
    }
    if (n & 1)
      cout << (p - g[n]) % p << endl;
    else
      cout << g[n] << endl;
    return 0;
    for (i = 1; i <= n; i++)
      for (g[i] = 0, j = 0; j < i; j++)
        g[i] += -g[j] * *min_element(a + j + 1, a + i + 1);
    for (i = 1; i <= n; i++) cout << g[i] << " \n"[i == n];
    if (n & 1)
      cout << -g[n] << endl;
    else
      cout << g[n] << endl;
  }
  