#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,A,B,i,j,ans,l,r,f[5010],mx,a[5010];
int main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	ans=1e18;
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);mx=0;
		f[i]=1e18;l=r=0;
		for(j=i-1;j;j--){
			if(a[j]<a[i]&&a[j]>mx)f[i]=min(f[i],f[j]+l*B+r*A);
			if(a[j]>a[i])r++;
			 else l++,mx=max(mx,a[j]);
		}
		if(mx==0)f[i]=r*A;
		for(j=i+1;j<=n;j++)mx=max(mx,a[j]);
		if(mx<a[i])ans=min(ans,f[i]+B*(n-i));
		//printf("%lld\n",f[i]);
	}
	printf("%lld",ans);
}
### Prompt

Generate 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<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,A,B,i,j,ans,l,r,f[5010],mx,a[5010];
int main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	ans=1e18;
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);mx=0;
		f[i]=1e18;l=r=0;
		for(j=i-1;j;j--){
			if(a[j]<a[i]&&a[j]>mx)f[i]=min(f[i],f[j]+l*B+r*A);
			if(a[j]>a[i])r++;
			 else l++,mx=max(mx,a[j]);
		}
		if(mx==0)f[i]=r*A;
		for(j=i+1;j<=n;j++)mx=max(mx,a[j]);
		if(mx<a[i])ans=min(ans,f[i]+B*(n-i));
		//printf("%lld\n",f[i]);
	}
	printf("%lld",ans);
}
```