#include<bits/stdc++.h>
using namespace std;

#define int long long
int dp[5003][5003] , arr[5003] , A , B , N;
void chkmin(int &a , int b){a = a < b ? a : b;}

signed main(){
	cin >> N >> A >> B;
	for(int i = 1 ; i <= N ; ++i) cin >> arr[i];
	memset(dp , 0x3f , sizeof(dp)); dp[0][0] = 0;
	for(int i = 0 ; i < N ; ++i)
		for(int j = 0 ; j <= N ; ++j)
			if(j < arr[i + 1]){
				chkmin(dp[i + 1][arr[i + 1]] , dp[i][j]);
				chkmin(dp[i + 1][j] , dp[i][j] + A);
			}
			else chkmin(dp[i + 1][j] , dp[i][j] + B);
	int mn = 1e18; for(int i = 1 ; i <= N ; ++i) mn = min(mn , dp[N][i]);
	cout << mn; return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:

* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.



Find the minimum total cost required to sort p in ascending order.

Constraints

* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.

Input

Input is given from Standard Input in the following format:


N A B
p_1 \cdots p_N


Output

Print the minimum total cost required to sort p in ascending order.

Examples

Input

3 20 30
3 1 2


Output

20


Input

4 20 30
4 2 3 1


Output

50


Input

1 10 10
1


Output

0


Input

4 1000000000 1000000000
4 3 2 1


Output

3000000000


Input

9 40 50
5 3 4 7 6 1 2 9 8


Output

220

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

#define int long long
int dp[5003][5003] , arr[5003] , A , B , N;
void chkmin(int &a , int b){a = a < b ? a : b;}

signed main(){
	cin >> N >> A >> B;
	for(int i = 1 ; i <= N ; ++i) cin >> arr[i];
	memset(dp , 0x3f , sizeof(dp)); dp[0][0] = 0;
	for(int i = 0 ; i < N ; ++i)
		for(int j = 0 ; j <= N ; ++j)
			if(j < arr[i + 1]){
				chkmin(dp[i + 1][arr[i + 1]] , dp[i][j]);
				chkmin(dp[i + 1][j] , dp[i][j] + A);
			}
			else chkmin(dp[i + 1][j] , dp[i][j] + B);
	int mn = 1e18; for(int i = 1 ; i <= N ; ++i) mn = min(mn , dp[N][i]);
	cout << mn; return 0;
}
```