#include <bits/stdc++.h>
using namespace std;
int n, top;
long long x, ans;
const int N = 2e5 + 5;
const int mo = 998244353;
long long a[N], f[N], sum[N], g[N], h[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  f[0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    while (a[i] <= a[g[top]]) top--;
    if (top)
      x = sum[i - 1] - sum[g[top] - 1];
    else
      x = sum[i - 1];
    x = -x * a[i] % mo;
    f[i] = (h[top] + x) % mo;
    sum[i] = (sum[i - 1] + f[i]) % mo;
    g[++top] = i;
    h[top] = (h[top - 1] + x) % mo;
  }
  ans = f[n];
  if (n % 2) ans = -f[n];
  ans = (ans % mo + mo) % mo;
  printf("%d\n", ans);
}

### Prompt

Create a solution in cpp for the following problem:

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>
using namespace std;
int n, top;
long long x, ans;
const int N = 2e5 + 5;
const int mo = 998244353;
long long a[N], f[N], sum[N], g[N], h[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  f[0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    while (a[i] <= a[g[top]]) top--;
    if (top)
      x = sum[i - 1] - sum[g[top] - 1];
    else
      x = sum[i - 1];
    x = -x * a[i] % mo;
    f[i] = (h[top] + x) % mo;
    sum[i] = (sum[i - 1] + f[i]) % mo;
    g[++top] = i;
    h[top] = (h[top - 1] + x) % mo;
  }
  ans = f[n];
  if (n % 2) ans = -f[n];
  ans = (ans % mo + mo) % mo;
  printf("%d\n", ans);
}

```