#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=5e3+10;
ll dp[N][N];
int P[N],D[N];
const ll INF=1e18;
int main()
{
	int n,A,B;scanf("%d%d%d",&n,&A,&B);
	for(int i=0;i<n;i++)scanf("%d",&D[i]);
	int tar=0;
	for(int i=0;i<n;i++){
		tar++;int cnt=0;
		for(int j=0;j<n;j++){
			if(D[j]==tar){
				P[tar]=n-tar-cnt;
				break;
			}else if(D[j]>tar)cnt++;
		}
	}fill(&dp[0][0],&dp[N-1][N],INF);
	for(int i=0;i<=n;i++)dp[0][i]=0;
	for(int i=1;i<=n;i++){
		ll M=INF;
		for(int j=0;j<=n;j++){
			M=min(M,dp[i-1][j]);
			//n-j,P[i]
			if(P[i]+1==n-j)dp[i][j+1]=min(dp[i][j+1],M);
			else if(P[i]+1<n-j)dp[i][j+1]=min(dp[i][j+1],M+B);
			else dp[i][j]=min(dp[i][j],M+A);
		}
	}cout<<dp[n][n]<<'\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>
using namespace std;
#define ll long long
const int N=5e3+10;
ll dp[N][N];
int P[N],D[N];
const ll INF=1e18;
int main()
{
	int n,A,B;scanf("%d%d%d",&n,&A,&B);
	for(int i=0;i<n;i++)scanf("%d",&D[i]);
	int tar=0;
	for(int i=0;i<n;i++){
		tar++;int cnt=0;
		for(int j=0;j<n;j++){
			if(D[j]==tar){
				P[tar]=n-tar-cnt;
				break;
			}else if(D[j]>tar)cnt++;
		}
	}fill(&dp[0][0],&dp[N-1][N],INF);
	for(int i=0;i<=n;i++)dp[0][i]=0;
	for(int i=1;i<=n;i++){
		ll M=INF;
		for(int j=0;j<=n;j++){
			M=min(M,dp[i-1][j]);
			//n-j,P[i]
			if(P[i]+1==n-j)dp[i][j+1]=min(dp[i][j+1],M);
			else if(P[i]+1<n-j)dp[i][j+1]=min(dp[i][j+1],M+B);
			else dp[i][j]=min(dp[i][j],M+A);
		}
	}cout<<dp[n][n]<<'\n';
}

```