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;
  const ll mod = 998244353;
  ll ad(ll a, ll b) { return (a + b + mod) % mod; }
  ll ml(ll a, ll b) { return ((a % mod) * (b % mod)) % mod; }
  const int N = 5e5;
  ll dp[N + 1][2] = {0};
  ll DP[N + 1] = {0};
  ll a[N + 1];
  ll dp_(int idx, int i) { return idx < 0 ? 0 : dp[idx][i]; }
  ll DP_(int idx) { return idx < 0 ? 0 : DP[idx]; }
  void solver() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    dp[0][1] = 0, dp[0][0] = 1, DP[0] = 1;
    ll total = 1;
    stack<int> s;
    for (int i = 1; i <= n; ++i) {
      total = ml(total, a[i]);
      dp[i][1] = ml(a[i], dp[i - 1][1]);
      while (!s.empty() && a[s.top()] > a[i]) s.pop();
      int idx = 0;
      if (!s.empty()) idx = s.top();
      if ((i - idx) % 2 == 0) {
        dp[i][1] = ad(dp[i][1], ml(a[i], DP_(idx - 1)));
        dp[i][1] = ad(dp[i][1], ml(a[i], DP_(i - 2)));
        dp[i][1] = ad(dp[i][1], dp[idx][1]);
        dp[i][1] = ad(dp[i][1], -ml(a[idx], dp_(idx - 1, 0)));
        dp[i][1] = ad(dp[i][1], -ml(a[idx], dp_(idx - 1, 1)));
      } else {
        dp[i][1] = ad(dp[i][1], ml(a[i], ad(DP_(i - 2), -DP_(idx - 1))));
        dp[i][1] = ad(dp[i][1], ad(ml(a[idx], dp_(idx - 1, 0)), -dp_(idx, 1)));
        dp[i][1] = ad(dp[i][1], ml(a[idx], dp_(idx - 1, 1)));
      }
      dp[i][0] = ad(total, -dp[i][1]);
      DP[i] = ad(dp[i][0], -DP[i - 1]);
      s.push(i);
    }
    cout << dp[n][0] << "\n";
    return;
  }
  int main() {
    solver();
    return 0;
  }
  