#include<bits/stdc++.h>
using namespace std;
int n,a,b,p[10005],i,j;
long long f[5005],s,ans=100000000000000ll;
int main()
{
	scanf("%d %d %d",&n,&a,&b);
	for(i=1;i<=n;++i)
	{
		scanf("%d",&p[i]);
		f[i]=100000000000000ll;
	}
	for(i=1;i<=n;++i)
	{
		s=0;
		for(j=i-1;j>=0;--j)
		{
			if(p[i]>p[j])
				f[i]=min(f[i],f[j]+s);
			if(p[j]>p[i])
				s+=a;
			else
				s+=b;
		}
	}
	for(i=n;i>=0;--i)
	{
		s=0;
		for(j=i+1;j<=n;++j)
			s+=b;
		ans=min(ans,f[i]+s);
	}
	cout<<ans;
}

### 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;
int n,a,b,p[10005],i,j;
long long f[5005],s,ans=100000000000000ll;
int main()
{
	scanf("%d %d %d",&n,&a,&b);
	for(i=1;i<=n;++i)
	{
		scanf("%d",&p[i]);
		f[i]=100000000000000ll;
	}
	for(i=1;i<=n;++i)
	{
		s=0;
		for(j=i-1;j>=0;--j)
		{
			if(p[i]>p[j])
				f[i]=min(f[i],f[j]+s);
			if(p[j]>p[i])
				s+=a;
			else
				s+=b;
		}
	}
	for(i=n;i>=0;--i)
	{
		s=0;
		for(j=i+1;j<=n;++j)
			s+=b;
		ans=min(ans,f[i]+s);
	}
	cout<<ans;
}

```