#include <bits/stdc++.h>
using namespace std;
const int MX = 200005, mod = 998244353;
int n;
long long a[MX], dp[MX];
stack<vector<long long> > st;
void main_() {
  cin >> n;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  long long acu = 0;
  dp[n] = 1;
  for (int i = n - 1; i >= 0; i--) {
    long long s = i % 2 ? -dp[i + 1] : dp[i + 1];
    while (st.size() && st.top()[0] >= a[i]) {
      (acu -= st.top()[0] * st.top()[1]) %= mod;
      (s += st.top()[1]) %= mod;
      st.pop();
    }
    (acu += a[i] * s) %= mod;
    st.push({a[i], s});
    dp[i] = i % 2 ? -acu : acu;
  }
  cout << ((((dp[0]) % (mod)) + (mod)) % (mod)) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) main_();
  return 0;
}

### Prompt

Generate 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 int MX = 200005, mod = 998244353;
int n;
long long a[MX], dp[MX];
stack<vector<long long> > st;
void main_() {
  cin >> n;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  long long acu = 0;
  dp[n] = 1;
  for (int i = n - 1; i >= 0; i--) {
    long long s = i % 2 ? -dp[i + 1] : dp[i + 1];
    while (st.size() && st.top()[0] >= a[i]) {
      (acu -= st.top()[0] * st.top()[1]) %= mod;
      (s += st.top()[1]) %= mod;
      st.pop();
    }
    (acu += a[i] * s) %= mod;
    st.push({a[i], s});
    dp[i] = i % 2 ? -acu : acu;
  }
  cout << ((((dp[0]) % (mod)) + (mod)) % (mod)) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) main_();
  return 0;
}

```