#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, N = 500005;
long long a, b, c, d, e, f, g = 1, h[N], arr[N], dp[N][2], pre[N][2];
string s;
vector<long long> v;
void solve() {
  cin >> a;
  for (long long i = 1; i <= a; i++) {
    cin >> arr[i];
  }
  stack<long long> stc;
  dp[0][0] = pre[0][0] = 1;
  for (long long i = 1; i <= a; i++) {
    while (stc.size() and arr[stc.top()] >= arr[i]) stc.pop();
    dp[i][0] = (!stc.size()
                    ? (pre[i - 1][1] * arr[i]) % mod
                    : (dp[stc.top()][0] +
                       (pre[i - 1][1] - pre[stc.top() - 1][1] + mod) * arr[i]) %
                          mod);
    dp[i][1] = (!stc.size()
                    ? (pre[i - 1][0] * arr[i]) % mod
                    : (dp[stc.top()][1] +
                       (pre[i - 1][0] - pre[stc.top() - 1][0] + mod) * arr[i]) %
                          mod);
    pre[i][0] = (dp[i][0] + pre[i - 1][0]) % mod;
    pre[i][1] = (dp[i][1] + pre[i - 1][1]) % mod;
    stc.push(i);
  }
  cout << (a % 2 ? (dp[a][1] - dp[a][0] + mod) % mod
                 : (dp[a][0] - dp[a][1] + mod) % mod)
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (g--) solve();
}

### 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 int mod = 998244353, N = 500005;
long long a, b, c, d, e, f, g = 1, h[N], arr[N], dp[N][2], pre[N][2];
string s;
vector<long long> v;
void solve() {
  cin >> a;
  for (long long i = 1; i <= a; i++) {
    cin >> arr[i];
  }
  stack<long long> stc;
  dp[0][0] = pre[0][0] = 1;
  for (long long i = 1; i <= a; i++) {
    while (stc.size() and arr[stc.top()] >= arr[i]) stc.pop();
    dp[i][0] = (!stc.size()
                    ? (pre[i - 1][1] * arr[i]) % mod
                    : (dp[stc.top()][0] +
                       (pre[i - 1][1] - pre[stc.top() - 1][1] + mod) * arr[i]) %
                          mod);
    dp[i][1] = (!stc.size()
                    ? (pre[i - 1][0] * arr[i]) % mod
                    : (dp[stc.top()][1] +
                       (pre[i - 1][0] - pre[stc.top() - 1][0] + mod) * arr[i]) %
                          mod);
    pre[i][0] = (dp[i][0] + pre[i - 1][0]) % mod;
    pre[i][1] = (dp[i][1] + pre[i - 1][1]) % mod;
    stc.push(i);
  }
  cout << (a % 2 ? (dp[a][1] - dp[a][0] + mod) % mod
                 : (dp[a][0] - dp[a][1] + mod) % mod)
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (g--) solve();
}

```