#include <bits/stdc++.h>
int main() {
  static int aa[200000], dp[200000 + 1], dq[200000], dr[200000], qu[200000];
  int n, i, cnt;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  dp[0] = n % 2 == 0 ? 1 : 998244353 - 1;
  cnt = 0;
  for (i = 0; i < n; i++) {
    int q = dp[i];
    while (cnt && aa[qu[cnt - 1]] >= aa[i]) {
      q = (q + dq[cnt - 1]) % 998244353;
      cnt--;
    }
    dq[cnt] = q, qu[cnt] = i,
    dr[cnt] = ((cnt == 0 ? 0 : dr[cnt - 1]) + (long long)q * aa[i]) % 998244353,
    cnt++;
    dp[i + 1] = (998244353 - dr[cnt - 1]) % 998244353;
  }
  printf("%d\n", dp[n]);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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].

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  static int aa[200000], dp[200000 + 1], dq[200000], dr[200000], qu[200000];
  int n, i, cnt;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  dp[0] = n % 2 == 0 ? 1 : 998244353 - 1;
  cnt = 0;
  for (i = 0; i < n; i++) {
    int q = dp[i];
    while (cnt && aa[qu[cnt - 1]] >= aa[i]) {
      q = (q + dq[cnt - 1]) % 998244353;
      cnt--;
    }
    dq[cnt] = q, qu[cnt] = i,
    dr[cnt] = ((cnt == 0 ? 0 : dr[cnt - 1]) + (long long)q * aa[i]) % 998244353,
    cnt++;
    dp[i + 1] = (998244353 - dr[cnt - 1]) % 998244353;
  }
  printf("%d\n", dp[n]);
  return 0;
}

```