#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int N = 500007;
stack<long long> s;
long long a[N];
long long f[N];
long long sum[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  s.push(0);
  f[0] = 1;
  sum[0] = 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    while (!s.empty() && a[s.top()] >= a[i]) {
      long long pos = s.top();
      s.pop();
      if (s.empty()) {
        ans = (ans - a[pos] * sum[pos - 1] % mod + mod) % mod;
      } else {
        ans = (ans - a[pos] * (sum[pos - 1] - sum[s.top() - 1] + mod) % mod +
               mod) %
              mod;
      }
    }
    if (s.empty()) {
      ans = (ans + a[i] * sum[i - 1] % mod) % mod;
    } else {
      ans = (ans + a[i] * ((sum[i - 1] - sum[s.top() - 1] + mod) % mod) % mod) %
            mod;
    }
    s.push(i);
    if (i % 2 == 1) {
      f[i] = ans;
    } else {
      f[i] = (mod - ans) % mod;
    }
    if (i % 2 == 1) {
      f[i] = (mod - f[i]) % mod;
    }
    sum[i] = (f[i] + sum[i - 1]) % mod;
  }
  if (n % 2 == 1) {
    printf("%lld\n", (mod - f[n]) % mod);
  } else {
    printf("%lld\n", f[n]);
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 500007;
stack<long long> s;
long long a[N];
long long f[N];
long long sum[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  s.push(0);
  f[0] = 1;
  sum[0] = 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    while (!s.empty() && a[s.top()] >= a[i]) {
      long long pos = s.top();
      s.pop();
      if (s.empty()) {
        ans = (ans - a[pos] * sum[pos - 1] % mod + mod) % mod;
      } else {
        ans = (ans - a[pos] * (sum[pos - 1] - sum[s.top() - 1] + mod) % mod +
               mod) %
              mod;
      }
    }
    if (s.empty()) {
      ans = (ans + a[i] * sum[i - 1] % mod) % mod;
    } else {
      ans = (ans + a[i] * ((sum[i - 1] - sum[s.top() - 1] + mod) % mod) % mod) %
            mod;
    }
    s.push(i);
    if (i % 2 == 1) {
      f[i] = ans;
    } else {
      f[i] = (mod - ans) % mod;
    }
    if (i % 2 == 1) {
      f[i] = (mod - f[i]) % mod;
    }
    sum[i] = (f[i] + sum[i - 1]) % mod;
  }
  if (n % 2 == 1) {
    printf("%lld\n", (mod - f[n]) % mod);
  } else {
    printf("%lld\n", f[n]);
  }
  return 0;
}

```