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;
  int a[500500], n;
  struct SegmentTree {
    long long w, val, lazy;
  } t[2000021];
  stack<int> mn;
  long long dp[500500];
  void pushdown(int x) {
    if (t[x].lazy) {
      t[x * 2].w = (t[x * 2].val * t[x].lazy) % 998244353;
      t[x * 2 + 1].w = (t[x * 2 + 1].val * t[x].lazy) % 998244353;
      t[x * 2].lazy = t[x].lazy, t[x * 2 + 1].lazy = t[x].lazy;
    }
  }
  void update(int x, int l, int r, int a, int b, long long k) {
    if (l >= a && r <= b) {
      t[x].lazy = k, t[x].w = (t[x].val * k) % 998244353;
      return;
    }
    int mid = (l + r) >> 1;
    if (mid >= a) update(x * 2, l, mid, a, b, k);
    if (mid < b) update(x * 2 + 1, mid + 1, r, a, b, k);
    t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
    t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
  }
  void insert(int x, int l, int r, int loc, long long k) {
    if (l == r) {
      t[x].val = k, t[x].w = (t[x].val * t[x].lazy) % 998244353;
      return;
    }
    pushdown(x);
    int mid = (l + r) >> 1;
    if (mid >= loc)
      insert(x * 2, l, mid, loc, k);
    else
      insert(x * 2 + 1, mid + 1, r, loc, k);
    t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
    t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = (1); i <= (n); i++) cin >> a[i];
    mn.push(0);
    dp[0] = 1;
    insert(1, 0, n, 0, 998244353 - dp[0]);
    for (int i = (1); i <= (n); i++) {
      while (a[mn.top()] > a[i]) mn.pop();
      update(1, 0, n, mn.top(), i - 1, a[i]);
      mn.push(i);
      dp[i] = ((i & 1) ? 998244353 - t[1].w : t[1].w) % 998244353;
      insert(1, 0, n, i, (i & 1) ? dp[i] : 998244353 - dp[i]);
    }
    cout << dp[n] << endl;
    return 0;
  }
  