#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int N = 500000;
int a[N + 5];
long long sum[N + 5], f[N + 5];
int main() {
  stack<array<int, 4>> s;
  s.push({0, 0, 0, 0});
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  long long tot = 0;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    int left = i;
    long long temp = 0;
    while (!s.empty() && s.top()[0] >= a[i]) {
      auto now = s.top();
      s.pop();
      tot = (tot - now[3] + mod) % mod;
      temp = (temp +
              (long long)a[i] * (sum[now[2]] - sum[now[1] - 1] + mod) % mod) %
             mod;
      left = now[1];
    }
    long long coef = i % 2 ? mod - 1 : 1;
    temp = (temp + (long long)a[i] * coef % mod * f[i - 1] % mod) % mod;
    s.push({a[i], left, i, (int)temp});
    tot = (tot + temp) % mod;
    f[i] = tot * coef % mod;
    sum[i] = (sum[i - 1] + f[i - 1] * coef % mod) % mod;
  }
  printf("%lld", f[n]);
  return 0;
}

### Prompt

Please create a solution in cpp 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 mod = 998244353;
const int N = 500000;
int a[N + 5];
long long sum[N + 5], f[N + 5];
int main() {
  stack<array<int, 4>> s;
  s.push({0, 0, 0, 0});
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  long long tot = 0;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    int left = i;
    long long temp = 0;
    while (!s.empty() && s.top()[0] >= a[i]) {
      auto now = s.top();
      s.pop();
      tot = (tot - now[3] + mod) % mod;
      temp = (temp +
              (long long)a[i] * (sum[now[2]] - sum[now[1] - 1] + mod) % mod) %
             mod;
      left = now[1];
    }
    long long coef = i % 2 ? mod - 1 : 1;
    temp = (temp + (long long)a[i] * coef % mod * f[i - 1] % mod) % mod;
    s.push({a[i], left, i, (int)temp});
    tot = (tot + temp) % mod;
    f[i] = tot * coef % mod;
    sum[i] = (sum[i - 1] + f[i - 1] * coef % mod) % mod;
  }
  printf("%lld", f[n]);
  return 0;
}

```