#include <bits/stdc++.h>
using namespace std;
int a[500500], n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  long long val = 0, sum = 0;
  stack<pair<long long, int> > s;
  for (int i = (1); i <= (n); i++) {
    val = sum + (i == 1);
    (sum += 998244353 - val * a[i] % 998244353) %= 998244353;
    while (!s.empty() && s.top().second > a[i]) {
      pair<long long, int> cur = s.top();
      s.pop();
      (sum += cur.first * (cur.second - a[i])) %= 998244353;
      (val += cur.first) %= 998244353;
    }
    s.push({val, a[i]});
  }
  cout << (n % 2 ? (998244353 - sum) % 998244353 : sum) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
int a[500500], n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  long long val = 0, sum = 0;
  stack<pair<long long, int> > s;
  for (int i = (1); i <= (n); i++) {
    val = sum + (i == 1);
    (sum += 998244353 - val * a[i] % 998244353) %= 998244353;
    while (!s.empty() && s.top().second > a[i]) {
      pair<long long, int> cur = s.top();
      s.pop();
      (sum += cur.first * (cur.second - a[i])) %= 998244353;
      (val += cur.first) %= 998244353;
    }
    s.push({val, a[i]});
  }
  cout << (n % 2 ? (998244353 - sum) % 998244353 : sum) << endl;
  return 0;
}

```