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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long MOD = 998244353;
  long long add(long long x, long long y) {
    x += y;
    if (x >= MOD) {
      x -= MOD;
    }
    return x;
  }
  long long sub(long long x, long long y) {
    x -= y;
    if (x < 0) {
      x += MOD;
    }
    return x;
  }
  long long mul(long long a, long long b) { return 1LL * a * b % MOD; }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long> ans(n + 1), pref(n + 1);
    ans[0] = pref[0] = 1;
    vector<long long> st;
    for (long long i = 1; i <= n; i++) {
      while (!st.empty() && a[st.back()] >= a[i - 1]) {
        st.pop_back();
      }
      long long kek = -1;
      if (!st.empty()) kek = st.back();
      st.push_back(i - 1);
      long long tmp = pref[i - 1];
      if (kek != -1) tmp = sub(tmp, pref[kek]);
      ans[i] = mul(a[i - 1], sub(0, tmp));
      if (kek != -1) {
        ans[i] = add(ans[i], ans[kek + 1]);
      }
      pref[i] = add(pref[i - 1], ans[i]);
    }
    if (n % 2) {
      ans[n] = sub(0, ans[n]);
    }
    cout << ans[n] << "\n";
  }
  