#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
typedef long long ll;
const ll inf=1e18;
ll dp[2][5555];int num[5555];
int main()
{
    int n,a,b;scanf("%d%d%d",&n,&a,&b);
    rep(i,1,n+1)scanf("%d",&num[i]);
    rep(i,1,n+1)dp[0][i]=inf;
    int type=1;dp[0][0]=0;
    rep(i,1,n+1)
    {
        rep(i,0,n+1)dp[type][i]=1e18;
        rep(j, 0, n+1)if(dp[!type][j]!=inf)
        {
            if(num[i]<j)dp[type][j]=min(dp[type][j],dp[!type][j]+b);
            else dp[type][j]=min(dp[type][j],dp[!type][j]+a),dp[type][num[i]]=min(dp[type][num[i]],dp[!type][j]);
        }
        type=!type;
    }
    ll res=inf;
    rep(i,1,n+1)res=min(res,dp[!type][i]);
    printf("%lld\n",res);
}
### Prompt

Your challenge is to write 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>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
typedef long long ll;
const ll inf=1e18;
ll dp[2][5555];int num[5555];
int main()
{
    int n,a,b;scanf("%d%d%d",&n,&a,&b);
    rep(i,1,n+1)scanf("%d",&num[i]);
    rep(i,1,n+1)dp[0][i]=inf;
    int type=1;dp[0][0]=0;
    rep(i,1,n+1)
    {
        rep(i,0,n+1)dp[type][i]=1e18;
        rep(j, 0, n+1)if(dp[!type][j]!=inf)
        {
            if(num[i]<j)dp[type][j]=min(dp[type][j],dp[!type][j]+b);
            else dp[type][j]=min(dp[type][j],dp[!type][j]+a),dp[type][num[i]]=min(dp[type][num[i]],dp[!type][j]);
        }
        type=!type;
    }
    ll res=inf;
    rep(i,1,n+1)res=min(res,dp[!type][i]);
    printf("%lld\n",res);
}
```