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
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,i,A,B,j,ans,f[5010][5010],a[5010];
int main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	memset(f,44,sizeof(f));
	f[0][0]=0;
	for(i=1;i<=n;i++){
		for(j=0;j<=n;j++){
			if(a[i]>j)f[i][a[i]]=min(f[i][a[i]],f[i-1][j]),f[i][j]=min(f[i][j],f[i-1][j]+A);
			 else f[i][j]=min(f[i][j],f[i-1][j]+B);
		}
	}
	ans=1e18;
	for(i=1;i<=n;i++)ans=min(ans,f[n][i]);
	printf("%lld",ans);
}
/*
枚举不动的数就行了 
*/