#include<bits/stdc++.h>
using namespace std;
const int N=5010;
typedef long long ll;
ll dp[N][N];
int f[N][N],n,A,B,a[N],b[N];
int main(){
	cin>>n>>A>>B;
	for(int i=1;i<=n;++i)cin>>a[i],b[a[i]]=i;
	for(int i=n;i;--i)
		for(int j=1;j<=n;++j)f[i][j]=f[i+1][j]+(a[i]>j);
	memset(dp,0x3f,sizeof dp);memset(dp[n+1],0,sizeof dp[n+1]);
	for(int i=n;i;--i){
		dp[i][n+1]=0;
		for(int j=n;j;--j)
			if(a[i]<j)dp[i][j]=dp[i+1][j];
			else if(b[j]<i)dp[i][j]=dp[i][j+1];
			else{
				int u=b[j];
				if(u==i)dp[i][j]=dp[i+1][j+1];
				else dp[i][j]=min(dp[i][j+1]+B,dp[u+1][j+1]+(ll)A*(f[i][j]-f[u][j]));
			}
	}
	cout<<dp[1][1]<<endl;
	return 0;
}


### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const int N=5010;
typedef long long ll;
ll dp[N][N];
int f[N][N],n,A,B,a[N],b[N];
int main(){
	cin>>n>>A>>B;
	for(int i=1;i<=n;++i)cin>>a[i],b[a[i]]=i;
	for(int i=n;i;--i)
		for(int j=1;j<=n;++j)f[i][j]=f[i+1][j]+(a[i]>j);
	memset(dp,0x3f,sizeof dp);memset(dp[n+1],0,sizeof dp[n+1]);
	for(int i=n;i;--i){
		dp[i][n+1]=0;
		for(int j=n;j;--j)
			if(a[i]<j)dp[i][j]=dp[i+1][j];
			else if(b[j]<i)dp[i][j]=dp[i][j+1];
			else{
				int u=b[j];
				if(u==i)dp[i][j]=dp[i+1][j+1];
				else dp[i][j]=min(dp[i][j+1]+B,dp[u+1][j+1]+(ll)A*(f[i][j]-f[u][j]));
			}
	}
	cout<<dp[1][1]<<endl;
	return 0;
}


```