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 p = 998244353;
  long long powmod(long long a, long long b) {
    b %= p - 1;
    long long r = 1;
    while (b) {
      if (b & 1) r = r * a % p;
      a = a * a % p;
      b >>= 1;
    }
    return r;
  }
  long long C(long long n, long long m) {
    if (n < m) return m;
    static vector<int> fac({1});
    for (int i = fac.size(); i < n + 1; i++)
      fac.push_back(1ll * i * fac.back() % p);
    return fac[n] * powmod(m, p - 2) % p * powmod(n - m, p - 2) % p;
  }
  int main() {
    int n;
    cin >> n;
    vector<long long> f(n + 1, 0), sf(n + 1, 0);
    auto sum = [&sf](const int &l, const int &r) {
      if (!l) return sf[r];
      return (sf[r] + p - sf[l - 1]) % p;
    };
    sf[0] = 1;
    auto add = [](long long &a, long long b) { a = (a + b) % p; };
    stack<array<int, 2> > st;
    st.push({0, 0});
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      while (!st.empty() && st.top()[0] > x) st.pop();
      int fa = st.top()[1];
      st.push({x, i});
      f[i] = (1ll * x * (i - 1 & 1 ? p - 1 : 1) % p * sum(fa, i - 1) % p +
              (i - fa & 1 ? p - 1ll : 1ll) * f[fa] % p) %
             p;
      sf[i] = (sf[i - 1] + (i & 1 ? p - 1ll : 1ll) * f[i]) % p;
    }
    cout << f[n] << endl;
    return 0;
  }
  