#include<bits/stdc++.h>
using namespace std;
int n,a,b,f[5005];
long long dp[5005][5005];
int main()
{
    scanf("%d%d%d",&n,&a,&b);
    for(int i=1;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        f[x]=i;
    }
    for(int i=1;i<=n;i++)
    {
        int pos=f[i];
        long long t=1e18;
        for(int j=1;j<=n;j++)
        {
            int p;
            if(pos==j) p=0;
            else if(pos>j) p=b;
            else p=a;
            t=min(t,dp[i-1][j]);
            dp[i][j]=t+p;
        }
    }
    long long ans=1e18;
    for(int i=1;i<=n;i++)
        ans=min(ans,dp[n][i]);
    printf("%lld\n",ans);
}

### Prompt

In cpp, your task is to solve 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;
int n,a,b,f[5005];
long long dp[5005][5005];
int main()
{
    scanf("%d%d%d",&n,&a,&b);
    for(int i=1;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        f[x]=i;
    }
    for(int i=1;i<=n;i++)
    {
        int pos=f[i];
        long long t=1e18;
        for(int j=1;j<=n;j++)
        {
            int p;
            if(pos==j) p=0;
            else if(pos>j) p=b;
            else p=a;
            t=min(t,dp[i-1][j]);
            dp[i][j]=t+p;
        }
    }
    long long ans=1e18;
    for(int i=1;i<=n;i++)
        ans=min(ans,dp[n][i]);
    printf("%lld\n",ans);
}

```