#include<bits/stdc++.h>
using namespace std ;
typedef long long ll ;
const ll N = 5055;
ll dp[N][N], A, B;
int n, cnt[N][N], p[N], pos[N]; 
int main()
{
	scanf("%d%lld%lld", &n, &A, &B) ;
	for(int i = 1 ; i <= n ; i ++){
		scanf("%d", &p[i]) ;
		pos[p[i]] = i ; 
	}
	for(int i = 2 ; i <= n ; i ++)
		for(int j = p[i]+1 ; j <= n ; j ++)	
			cnt[i][j] = cnt[i][j-1] + (pos[j] < i) ;
	for(int i = 1 ; i <= n ; i ++)
		for(int j = 1 ; j <= n ; j ++)
			if(p[i] > j) dp[i][j] = dp[i-1][j] ;
			else dp[i][j] = min(dp[i-1][j]+B, cnt[i][j]*A+dp[i-1][p[i]-1]) ; 
	cout << dp[n][n] ;
}
### Prompt

Create a solution in Cpp for the following problem:

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 ;
typedef long long ll ;
const ll N = 5055;
ll dp[N][N], A, B;
int n, cnt[N][N], p[N], pos[N]; 
int main()
{
	scanf("%d%lld%lld", &n, &A, &B) ;
	for(int i = 1 ; i <= n ; i ++){
		scanf("%d", &p[i]) ;
		pos[p[i]] = i ; 
	}
	for(int i = 2 ; i <= n ; i ++)
		for(int j = p[i]+1 ; j <= n ; j ++)	
			cnt[i][j] = cnt[i][j-1] + (pos[j] < i) ;
	for(int i = 1 ; i <= n ; i ++)
		for(int j = 1 ; j <= n ; j ++)
			if(p[i] > j) dp[i][j] = dp[i-1][j] ;
			else dp[i][j] = min(dp[i-1][j]+B, cnt[i][j]*A+dp[i-1][p[i]-1]) ; 
	cout << dp[n][n] ;
}
```