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 long long inf = 1e9 + 7;
  const long long longinf = 1LL << 60;
  const long long mod = 998244353;
  const long long N = 5e5 + 10;
  long long a[N];
  long long dp[N][2];
  signed main() {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    stack<pair<long long, long long>> even, odd;
    long long esum = a[1], osum = 0;
    even.emplace(a[1], 1);
    odd.emplace(a[1], 0);
    for (long long i = 1; i < n; i++) {
      dp[i + 1][0] = osum;
      dp[i + 1][1] = esum;
      long long os = 0;
      while (odd.size() && odd.top().first > a[i + 1]) {
        osum += mod - odd.top().first * odd.top().second % mod;
        if (osum >= mod) osum -= mod;
        os += odd.top().second;
        os %= mod;
        odd.pop();
      }
      osum += a[i + 1] * (os + dp[i + 1][1]) % mod;
      osum %= mod;
      long long es = 0;
      while (even.size() && even.top().first > a[i + 1]) {
        esum += mod - even.top().first * even.top().second % mod;
        if (esum >= mod) esum -= mod;
        es += even.top().second;
        es %= mod;
        even.pop();
      }
      esum += a[i + 1] * (es + dp[i + 1][0]) % mod;
      esum %= mod;
      odd.emplace(a[i + 1], (os + dp[i + 1][1] % mod));
      even.emplace(a[i + 1], (es + dp[i + 1][0]) % mod);
    }
    if (n % 2)
      cout << (esum - osum + mod) % mod << endl;
    else
      cout << (osum - esum + mod) % mod << endl;
    return 0;
  }
  