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 int read() {
    int x = 0;
    char ch = getchar();
    bool f = 0;
    for (; ch < '0' || ch > '9'; ch = getchar())
      if (ch == '-') f = 1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
      x = (x << 3) + (x << 1) + (ch ^ 48);
    return f ? -x : x;
  }
  const long long mod = 998244353;
  inline long long Add(long long x, long long y) {
    x += y;
    return (x < mod) ? x : x - mod;
  }
  inline long long Dec(long long x, long long y) {
    x -= y;
    return (x < 0) ? x + mod : x;
  }
  inline long long Mul(long long x, long long y) { return x * y % mod; }
  inline long long Pow(long long x, long long y) {
    y %= (mod - 1);
    long long ans = 1;
    for (; y; y >>= 1, x = x * x % mod)
      if (y & 1) ans = ans * x % mod;
    return ans;
  }
  const int N = 5e5 + 5;
  int n, a[N];
  long long s[N], f[N];
  int st[N], top;
  int main() {
    n = read();
    for (int i = (1), end_i = (n); i <= end_i; i++) a[i] = read();
    s[0] = 1;
    f[0] = 1;
    long long now = 0;
    for (int i = (1), end_i = (n); i <= end_i; i++) {
      for (; top && a[st[top]] > a[i]; top--)
        now = Dec(now, Dec(s[st[top] - 1], st[top - 1] ? s[st[top - 1] - 1] : 0) *
                           a[st[top]] % mod);
      now = Add(now, Dec(s[i - 1], st[top] ? s[st[top] - 1] : 0) * a[i] % mod);
      st[++top] = i;
      f[i] = (mod - now) % mod;
      s[i] = Add(s[i - 1], f[i]);
    }
    printf("%lld\n", f[n] * Pow(mod - 1, n) % mod);
    return 0;
  }
  