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;
  void main0();
  int main() {
    clock_t start, end;
    ios::sync_with_stdio(false);
    cin.tie(0);
    main0();
    return 0;
  }
  const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
  const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  const int N = 1e6 + 5;
  const int M = 1e5;
  const int INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3f;
  const int mod = 998244353;
  const double eps = 1e-6;
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  bool chkmin(T &x, T y) {
    return y < x ? x = y, 1 : 0;
  }
  template <typename T>
  bool chkmax(T &x, T y) {
    return y > x ? x = y, 1 : 0;
  }
  int n;
  long long a[N];
  int s[N], p = 0;
  long long dp[N];
  long long pre[N];
  long long query(int l, int r) { return pre[r] - (l ? pre[l - 1] : 0); }
  void main0() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    dp[0] = pre[0] = 1;
    long long sum = 0;
    for (int i = 1; i <= n; ++i) {
      while (p && (a[s[p]] >= a[i])) {
        sum -= a[s[p]] * query(s[p - 1], s[p] - 1);
        sum %= mod;
        p--;
      }
      s[++p] = i;
      sum += a[i] * query(s[p - 1], s[p] - 1);
      sum %= mod;
      dp[i] = (-sum + mod) % mod;
      pre[i] = (pre[i - 1] + dp[i]) % mod;
    }
    long long ans = dp[n] * (n & 1 ? -1 : 1);
    ans = (ans + mod) % mod;
    cout << ans << endl;
  }
  