#include<stdio.h>
#include<string.h>
typedef long long ll;
const ll inf=1e18;
void fmin(ll&a,ll b){
	b<a?a=b:0;
}
int p[5010];
ll f[5010],g[5010],tf[5010],tg[5010];
int main(){
	int n,a,b,i,j;
	ll t;
	scanf("%d%d%d",&n,&a,&b);
	for(i=1;i<=n;i++){
		scanf("%d",&j);
		p[j]=i;
	}
	for(i=1;i<=n;i++){
		t=inf;
		for(j=1;j<=n;j++){
			fmin(t,g[j-1]);
			tf[j]=t+(p[i]<j?a:(p[i]>j?b:0));
			fmin(t,f[j]);
		}
		t=inf;
		for(j=0;j<=n;j++){
			fmin(t,g[j]);
			tg[j]=t+(p[i]<=j?a:(p[i]>j?b:0));
			fmin(t,f[j+1]);
		}
		memcpy(f,tf,sizeof(tf));
		memcpy(g,tg,sizeof(tg));
	}
	t=inf;
	for(i=1;i<=n;i++)fmin(t,f[i]);
	for(i=0;i<=n;i++)fmin(t,g[i]);
	printf("%lld",t);
}
### Prompt

Please formulate a Cpp solution 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<stdio.h>
#include<string.h>
typedef long long ll;
const ll inf=1e18;
void fmin(ll&a,ll b){
	b<a?a=b:0;
}
int p[5010];
ll f[5010],g[5010],tf[5010],tg[5010];
int main(){
	int n,a,b,i,j;
	ll t;
	scanf("%d%d%d",&n,&a,&b);
	for(i=1;i<=n;i++){
		scanf("%d",&j);
		p[j]=i;
	}
	for(i=1;i<=n;i++){
		t=inf;
		for(j=1;j<=n;j++){
			fmin(t,g[j-1]);
			tf[j]=t+(p[i]<j?a:(p[i]>j?b:0));
			fmin(t,f[j]);
		}
		t=inf;
		for(j=0;j<=n;j++){
			fmin(t,g[j]);
			tg[j]=t+(p[i]<=j?a:(p[i]>j?b:0));
			fmin(t,f[j+1]);
		}
		memcpy(f,tf,sizeof(tf));
		memcpy(g,tg,sizeof(tg));
	}
	t=inf;
	for(i=1;i<=n;i++)fmin(t,f[i]);
	for(i=0;i<=n;i++)fmin(t,g[i]);
	printf("%lld",t);
}
```