#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=5010;
ll f[N][N<<1];
int n,a[N],x;
ll A,B;
ll pan(int x,int y)
{
	if (x<y) return B;
	else if (x>y) return A;
	else return 0;
}
ll pan1(int x,int y)
{
	if (x-1>=y) return A;
	else return B;
}
int main() {
	scanf("%d",&n);
	scanf("%lld%lld",&A,&B);
	for (int i=1;i<=n;i++) scanf("%d",&x),a[x]=i;
	for (int i=1;i<=n;i++) {
		for (int j=1;j<=n*2+1;j++) 
		 if (j%2==0) f[i][j]=f[i-1][j-1]+pan(j/2,a[i]);
		 else f[i][j]=f[i-1][j]+pan1(j/2+1,a[i]);
		for (int j=2;j<=n*2+1;j++) f[i][j]=min(f[i][j],f[i][j-1]); 
	}
	ll ans=10000000000000000;
	for (int i=1;i<=n*2+1;i++) ans=min(ans,f[n][i]);
	cout << ans << endl;
} 
### 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;
const int N=5010;
ll f[N][N<<1];
int n,a[N],x;
ll A,B;
ll pan(int x,int y)
{
	if (x<y) return B;
	else if (x>y) return A;
	else return 0;
}
ll pan1(int x,int y)
{
	if (x-1>=y) return A;
	else return B;
}
int main() {
	scanf("%d",&n);
	scanf("%lld%lld",&A,&B);
	for (int i=1;i<=n;i++) scanf("%d",&x),a[x]=i;
	for (int i=1;i<=n;i++) {
		for (int j=1;j<=n*2+1;j++) 
		 if (j%2==0) f[i][j]=f[i-1][j-1]+pan(j/2,a[i]);
		 else f[i][j]=f[i-1][j]+pan1(j/2+1,a[i]);
		for (int j=2;j<=n*2+1;j++) f[i][j]=min(f[i][j],f[i][j-1]); 
	}
	ll ans=10000000000000000;
	for (int i=1;i<=n*2+1;i++) ans=min(ans,f[n][i]);
	cout << ans << endl;
} 
```