#include<bits/stdc++.h>
#define ll long long
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
using namespace std;
const int N=5005;
int n,A,B,a[N];
int p[N];
ll f[2][N];
int main(){
	scanf("%d%d%d",&n,&A,&B);
	For(i,1,n) scanf("%d",&a[i]),p[a[i]]=i;
	For(i,1,n) For(j,0,n){
		int co=(j==p[i]?0:(j<p[i]?B:A));
		f[i&1][j]=min(j==0?1ll<<60:f[i&1][j-1],f[(i&1)^1][j]+co);
	}
	printf("%lld\n",f[n&1][n]);
}
### Prompt

Develop a solution in cpp to the problem described below:

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
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
using namespace std;
const int N=5005;
int n,A,B,a[N];
int p[N];
ll f[2][N];
int main(){
	scanf("%d%d%d",&n,&A,&B);
	For(i,1,n) scanf("%d",&a[i]),p[a[i]]=i;
	For(i,1,n) For(j,0,n){
		int co=(j==p[i]?0:(j<p[i]?B:A));
		f[i&1][j]=min(j==0?1ll<<60:f[i&1][j-1],f[(i&1)^1][j]+co);
	}
	printf("%lld\n",f[n&1][n]);
}
```