#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
int main() {
  int n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<long long> dp(n + 2);
  vector<long long> sum(n + 2);
  vector<long long> ndp(n + 2);
  sum[0] = dp[0] = 0;
  sum[1] = dp[1] = 1;
  stack<int> st;
  st.push(0);
  a[0] = 0;
  for (int i = 1, idx; i <= n; i++) {
    idx = st.top();
    while (a[idx] >= a[i]) {
      st.pop();
      idx = st.top();
    }
    st.push(i);
    i++;
    idx++;
    ndp[i] = (ndp[idx] + (sum[i - 1] - sum[idx - 1]) * a[i - 1]) % mod;
    dp[i] = (i % 2 == 1 ? -1 : 1) * ndp[i];
    sum[i] = (sum[i - 1] + (i % 2 == 1 ? 1 : -1) * dp[i]) % mod;
    i--;
  }
  cout << (dp[n + 1] + mod) % mod << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int main() {
  int n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<long long> dp(n + 2);
  vector<long long> sum(n + 2);
  vector<long long> ndp(n + 2);
  sum[0] = dp[0] = 0;
  sum[1] = dp[1] = 1;
  stack<int> st;
  st.push(0);
  a[0] = 0;
  for (int i = 1, idx; i <= n; i++) {
    idx = st.top();
    while (a[idx] >= a[i]) {
      st.pop();
      idx = st.top();
    }
    st.push(i);
    i++;
    idx++;
    ndp[i] = (ndp[idx] + (sum[i - 1] - sum[idx - 1]) * a[i - 1]) % mod;
    dp[i] = (i % 2 == 1 ? -1 : 1) * ndp[i];
    sum[i] = (sum[i - 1] + (i % 2 == 1 ? 1 : -1) * dp[i]) % mod;
    i--;
  }
  cout << (dp[n + 1] + mod) % mod << '\n';
  return 0;
}

```