#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5005;
int n,a,b,i,j,p[N],s[N][N],x,y;
ll f[N],ans;
int ask(int a,int b,int c,int d){return s[b][d]-s[b][c-1]-s[a-1][d]+s[a-1][c-1];}
int main(){
	scanf("%d%d%d",&n,&a,&b);ans=1ll*n*min(a,b);
	for(i=1;i<=n;++i)for(scanf("%d",p+i),memcpy(s[i]+1,s[i-1]+1,n<<2),j=p[i];j<=n;++j)++s[i][j];
	for(i=1;i<=n;++i){
		x=ask(1,i-1,p[i]+1,n);f[i]=1ll*x*a+1ll*(i-1-x)*min(a,b);
		for(j=1;j<i;++j)if(p[j]<p[i]){
			x=ask(j+1,i-1,1,p[j]-1);y=ask(j+1,i-1,p[i]+1,n);
			f[i]=min(f[i],1ll*x*b+1ll*y*a+1ll*(i-j-1-x-y)*min(a,b)+f[j]);
		}
		x=ask(i+1,n,1,p[i]-1);ans=min(ans,f[i]+1ll*x*b+1ll*(n-i-x)*min(a,b));
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

In CPP, your task is to solve 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>
using namespace std;
typedef long long ll;
const int N=5005;
int n,a,b,i,j,p[N],s[N][N],x,y;
ll f[N],ans;
int ask(int a,int b,int c,int d){return s[b][d]-s[b][c-1]-s[a-1][d]+s[a-1][c-1];}
int main(){
	scanf("%d%d%d",&n,&a,&b);ans=1ll*n*min(a,b);
	for(i=1;i<=n;++i)for(scanf("%d",p+i),memcpy(s[i]+1,s[i-1]+1,n<<2),j=p[i];j<=n;++j)++s[i][j];
	for(i=1;i<=n;++i){
		x=ask(1,i-1,p[i]+1,n);f[i]=1ll*x*a+1ll*(i-1-x)*min(a,b);
		for(j=1;j<i;++j)if(p[j]<p[i]){
			x=ask(j+1,i-1,1,p[j]-1);y=ask(j+1,i-1,p[i]+1,n);
			f[i]=min(f[i],1ll*x*b+1ll*y*a+1ll*(i-j-1-x-y)*min(a,b)+f[j]);
		}
		x=ask(i+1,n,1,p[i]-1);ans=min(ans,f[i]+1ll*x*b+1ll*(n-i-x)*min(a,b));
	}
	printf("%lld\n",ans);
	return 0;
}
```