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 std::cin;
  using std::cout;
  using std::endl;
  using std::string;
  const int kN = 2e5 + 5;
  const int kInf = 1e9 + 7;
  const int P = 998244353;
  int n, a[kN], lm[kN];
  long long f[kN], s1[kN], s2[kN];
  int sgn(int x) { return (x & 1) ? P - 1 : 1; }
  void solve(int l, int r) {
    if (l == r) return;
    int mid = (l + r) >> 1;
    solve(l, mid);
    int minv = kInf;
    for (int i = mid; i >= l; --i) {
      minv = std::min(minv, a[i + 1]);
      s1[i] = f[i] * sgn(mid - i) % P * minv % P;
      s2[i] = f[i] * sgn(mid - i) % P;
      if (i < mid) {
        (s1[i] += s1[i + 1]) %= P;
        (s2[i] += s2[i + 1]) %= P;
      }
    }
    minv = kInf;
    int p = 0, lb = 0;
    for (int i = mid + 1; i <= r; ++i) {
      if (a[i] < minv) {
        minv = a[i];
        p = i;
      }
      lb = std::max(lm[p], l + 1);
      if (lb <= mid + 1) {
        (f[i] += s2[lb - 1] * minv % P * sgn(i - mid + 1)) %= P;
      }
      (f[i] += (s1[l] - s1[lb - 1] + P) * sgn(i - mid + 1) % P) %= P;
    }
    solve(mid + 1, r);
  }
  int main() {
    std::ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[lm[i] = i];
    for (int i = 1; i <= n; ++i) {
      while (lm[i] > 1 && a[lm[i] - 1] >= a[i]) lm[i] = lm[lm[i] - 1];
    }
    f[0] = 1;
    solve(0, n);
    printf("%lld", f[n]);
    return 0;
  }
  