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;
  using ll = long long;
  constexpr int mod = 998244353;
  ll _inv(ll a, ll m = mod) {
    ll b = m, u = 1, v = 0;
    while (b) {
      ll t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
  }
  int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<vector<ll>> dp(n + 1, vector<ll>(2, 0));
    vector<ll> sum(2);
    vector<stack<pair<int, ll>>> st(2);
    dp[0][0] = 1;
    for (int i = 0; i < n; i++) {
      vector<ll> tmp(2);
      for (int j = 0; j < 2; j++) {
        while (!st[j].empty() && st[j].top().first > a[i]) {
          (sum[j] += mod - st[j].top().second) %= mod;
          (tmp[j] += st[j].top().second * a[i] % mod * _inv(st[j].top().first) %
                     mod) %= mod;
          st[j].pop();
        }
        dp[i + 1][1 - j] = (tmp[j] + sum[j] + dp[i][1 - j] * a[i]) % mod;
        st[j].push({a[i], (tmp[j] + dp[i][1 - j] * a[i]) % mod});
        (sum[j] += st[j].top().second) %= mod;
      }
      swap(st[0], st[1]), swap(sum[0], sum[1]);
    }
    cout << (dp[n][0] - dp[n][1] + mod) % mod << endl;
  }
  