#include <bits/stdc++.h>
using namespace std;
const long long dim = 500009, mod = 998244353;
struct elem {
  long long x, s;
} stiva[dim];
long long vf = 1;
long long n, a[dim];
long long dp[dim], sum[dim];
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  dp[0] = 1;
  sum[1] = a[1] % mod;
  stiva[vf] = {a[vf], vf};
  for (long long i = 1; i <= n; i++) {
    long long x = a[i + 1], s;
    if (i % 2 == 1) {
      dp[i] = sum[vf];
      s = (mod - dp[i]) % mod;
    } else {
      dp[i] = (mod - sum[vf]) % mod;
      s = dp[i];
    }
    while (vf && stiva[vf].x > x) {
      s += stiva[vf--].s;
      s %= mod;
    }
    stiva[++vf] = {x, s};
    sum[vf] = sum[vf - 1] + x * s;
    sum[vf] %= mod;
  }
  cout << dp[n];
}

### Prompt

Your task is to create 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>
using namespace std;
const long long dim = 500009, mod = 998244353;
struct elem {
  long long x, s;
} stiva[dim];
long long vf = 1;
long long n, a[dim];
long long dp[dim], sum[dim];
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  dp[0] = 1;
  sum[1] = a[1] % mod;
  stiva[vf] = {a[vf], vf};
  for (long long i = 1; i <= n; i++) {
    long long x = a[i + 1], s;
    if (i % 2 == 1) {
      dp[i] = sum[vf];
      s = (mod - dp[i]) % mod;
    } else {
      dp[i] = (mod - sum[vf]) % mod;
      s = dp[i];
    }
    while (vf && stiva[vf].x > x) {
      s += stiva[vf--].s;
      s %= mod;
    }
    stiva[++vf] = {x, s};
    sum[vf] = sum[vf - 1] + x * s;
    sum[vf] %= mod;
  }
  cout << dp[n];
}

```