#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,A,B,a[5030];
long long f[5030][5030];
int main()
{
	scanf("%d%d%d",&n,&A,&B);
	memset(f,0x3f,sizeof(f));
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	f[0][0]=0;
	for(int i=1;i<=n;i++)
		for(int j=i-1;j>=0;j--)
			{
				if(a[i]<a[j]) f[i][j]=min(f[i-1][j]+B,f[i][j]);
				else f[i][j]=min(f[i-1][j]+A,f[i][j]),f[i][i]=min(f[i][i],f[i-1][j]);
			}
	long long ans=f[0][1];
	for(int i=1;i<=n;i++) ans=min(ans,f[n][i]);
	printf("%lld\n",ans);
	return 0;
}	
### Prompt

Please create a solution in CPP to 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<cstdio>
#include<cstring>
using namespace std;
int n,A,B,a[5030];
long long f[5030][5030];
int main()
{
	scanf("%d%d%d",&n,&A,&B);
	memset(f,0x3f,sizeof(f));
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	f[0][0]=0;
	for(int i=1;i<=n;i++)
		for(int j=i-1;j>=0;j--)
			{
				if(a[i]<a[j]) f[i][j]=min(f[i-1][j]+B,f[i][j]);
				else f[i][j]=min(f[i-1][j]+A,f[i][j]),f[i][i]=min(f[i][i],f[i-1][j]);
			}
	long long ans=f[0][1];
	for(int i=1;i<=n;i++) ans=min(ans,f[n][i]);
	printf("%lld\n",ans);
	return 0;
}	
```