#include <bits/stdc++.h>
const int maxn = 200005, mod = 998244353;
int n, top, ans;
int a[maxn], stk[maxn][2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int val = (ans + (i == 1)) % mod;
    ans = (ans - 1ll * val * a[i] % mod + mod) % mod;
    while (top > 0 && stk[top][1] > a[i])
      ans = (ans + 1ll * stk[top][0] * (stk[top][1] - a[i])) % mod,
      val = (val + stk[top][0]) % mod, top--;
    stk[++top][0] = val, stk[top][1] = a[i];
  }
  printf("%d\n", n % 2 == 0 ? ans : (mod - ans) % mod);
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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>
const int maxn = 200005, mod = 998244353;
int n, top, ans;
int a[maxn], stk[maxn][2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int val = (ans + (i == 1)) % mod;
    ans = (ans - 1ll * val * a[i] % mod + mod) % mod;
    while (top > 0 && stk[top][1] > a[i])
      ans = (ans + 1ll * stk[top][0] * (stk[top][1] - a[i])) % mod,
      val = (val + stk[top][0]) % mod, top--;
    stk[++top][0] = val, stk[top][1] = a[i];
  }
  printf("%d\n", n % 2 == 0 ? ans : (mod - ans) % mod);
  return 0;
}

```