#include<bits/stdc++.h>
using namespace std;
#define int long long 
const int N=5005;
int n,A,B,dp[N][N*2],g[N][N*2],p[N],q[N];
signed main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	for (int i=1;i<=n;i++)scanf("%lld",&p[i]);
	for (int i=1;i<=n;i++)q[p[i]]=i;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=2*n+1;j++){
			if (j%2==0)dp[i][j]=g[i-1][j-1];
			else dp[i][j]=g[i-1][j];
			if (j<q[i]*2)dp[i][j]+=B;
			else if (j>q[i]*2)dp[i][j]+=A;
			g[i][j]=min(dp[i][j],(j==1?1000000000000000000ll:g[i][j-1]));
		}
	printf("%lld\n",g[n][2*n+1]);
}
### Prompt

Create a solution in CPP for 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;
#define int long long 
const int N=5005;
int n,A,B,dp[N][N*2],g[N][N*2],p[N],q[N];
signed main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	for (int i=1;i<=n;i++)scanf("%lld",&p[i]);
	for (int i=1;i<=n;i++)q[p[i]]=i;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=2*n+1;j++){
			if (j%2==0)dp[i][j]=g[i-1][j-1];
			else dp[i][j]=g[i-1][j];
			if (j<q[i]*2)dp[i][j]+=B;
			else if (j>q[i]*2)dp[i][j]+=A;
			g[i][j]=min(dp[i][j],(j==1?1000000000000000000ll:g[i][j-1]));
		}
	printf("%lld\n",g[n][2*n+1]);
}
```