#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int pos[5010];
long long f[2][5010];
int n;
int a,b;
int main()
{
	scanf("%d %d %d",&n,&a,&b);
	for(int i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		pos[x]=i;
	}
	for(int i=1;i<=n;i++)
		if(i<pos[1])f[1][i]=b;
		else if(i>pos[1])f[1][i]=a;
		else f[1][i]=0;
	for(int i=2;i<=n;i++)
	{
		int now=i&1;
		int lst=now^1;
		for(int j=1;j<=n;j++)
			if(j<pos[i])
				f[now][j]=f[lst][j]+b;
			else if(j==pos[i])
			{
				f[now][j]=1e18;
				for(int k=1;k<=j;k++)
					f[now][j]=min(f[now][j],f[lst][k]);
			}
			else f[now][j]=f[lst][j]+a;
	}
	long long ans=1e18;
	for(int i=1;i<=n;i++)
		ans=min(ans,f[n&1][i]);
	printf("%lld\n",ans);
	return 0;
}
### 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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int pos[5010];
long long f[2][5010];
int n;
int a,b;
int main()
{
	scanf("%d %d %d",&n,&a,&b);
	for(int i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		pos[x]=i;
	}
	for(int i=1;i<=n;i++)
		if(i<pos[1])f[1][i]=b;
		else if(i>pos[1])f[1][i]=a;
		else f[1][i]=0;
	for(int i=2;i<=n;i++)
	{
		int now=i&1;
		int lst=now^1;
		for(int j=1;j<=n;j++)
			if(j<pos[i])
				f[now][j]=f[lst][j]+b;
			else if(j==pos[i])
			{
				f[now][j]=1e18;
				for(int k=1;k<=j;k++)
					f[now][j]=min(f[now][j],f[lst][k]);
			}
			else f[now][j]=f[lst][j]+a;
	}
	long long ans=1e18;
	for(int i=1;i<=n;i++)
		ans=min(ans,f[n&1][i]);
	printf("%lld\n",ans);
	return 0;
}
```