#include<bits/stdc++.h>
const int N=5005;
using namespace std;
typedef long long ll;
const ll inf=1e18;
int n,a,b,p[N],x;
ll f[N],g[N],ans=inf;
int main(){
	scanf("%d%d%d",&n,&a,&b);
	for(int i=0;i<n;i++)
	scanf("%d",&x),p[x-1]=i;
	for(int i=0;i<n;i++){
		for(int j=0;j<=n;j++)
		g[j]=f[j],f[j]=inf;
		for(int j=0;j<=n;j++)
		if(p[i]<j)
		f[j]=min(f[j],g[j]+a);
		else{
			if(p[i]>j)f[j]=min(f[j],g[j]+b);
			f[p[i]+1]=min(f[p[i]+1],g[j]);
		}
	}
	for(int i=0;i<=n;i++)
	ans=min(ans,f[i]);
	printf("%lld",ans);
}
### 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<bits/stdc++.h>
const int N=5005;
using namespace std;
typedef long long ll;
const ll inf=1e18;
int n,a,b,p[N],x;
ll f[N],g[N],ans=inf;
int main(){
	scanf("%d%d%d",&n,&a,&b);
	for(int i=0;i<n;i++)
	scanf("%d",&x),p[x-1]=i;
	for(int i=0;i<n;i++){
		for(int j=0;j<=n;j++)
		g[j]=f[j],f[j]=inf;
		for(int j=0;j<=n;j++)
		if(p[i]<j)
		f[j]=min(f[j],g[j]+a);
		else{
			if(p[i]>j)f[j]=min(f[j],g[j]+b);
			f[p[i]+1]=min(f[p[i]+1],g[j]);
		}
	}
	for(int i=0;i<=n;i++)
	ans=min(ans,f[i]);
	printf("%lld",ans);
}
```