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;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') f = ch == '-' ? -1 : 1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
    return x * f;
  }
  const long long N(2e5 + 10);
  const long long mod(998244353);
  long long n;
  long long a[N];
  long long g[N];
  long long s[N];
  long long pre[N];
  long long nxt[N];
  long long stk[N];
  signed main() {
    n = read();
    a[1] = a[n + 2] = -1e9;
    long long top = 1;
    stk[1] = 1;
    for (long long i = (2); i <= (n + 1); i++) a[i] = read();
    for (long long i = (2); i <= (n + 2); i++) {
      for (; top && a[stk[top]] >= a[i]; top--) nxt[stk[top]] = i;
      pre[i] = stk[top], stk[++top] = i;
    }
    s[1] = 1;
    for (long long i = (2); i <= (n + 1); i++) {
      long long x = s[i - 1] - s[pre[i] - 1] + mod;
      x = (x % mod * a[i]) % mod, (g[i] += g[i - 1] + x) %= mod;
      (g[nxt[i]] += mod - x) %= mod, s[i] = (s[i - 1] + mod - g[i]) % mod;
    }
    if ((n + 1) & 1) g[n + 1] = mod - g[n + 1];
    cout << g[n + 1] % mod;
    return 0;
  }
  