#include <iostream>
using namespace std;
typedef long long LL;
const int N=5022;
LL dp[N][N];
int n,a,b,x[N],pos[N];
int main(){
    scanf("%d%d%d",&n,&a,&b);
    for(int i=1;i<=n;i++) scanf("%d",&x[i]), pos[x[i]]=i;
    for(int i=0;i<N;i++)for(int j=0;j<N;j++)dp[i][j]=1e18;
    dp[0][0]=0;
    for(int i=1;i<=n;i++){
        LL mn=dp[i-1][0];
        for(int j=0;j<=n+1;j++){
            mn=min(mn,dp[i-1][j]);

            if(j==pos[i]) dp[i][j]=mn;
            else if(j<pos[i]) dp[i][j]=mn+min(1LL*b,1LL*a*(pos[i]-j));
            else dp[i][j]=mn+min(1LL*a,1LL*b*(j-pos[i]));
    
            //printf("dp[%d][%d]=%lld\n", i,j,dp[i][j]);
        }
    }
    LL ans=1e18;
    for(int i=0;i<=n+1;i++)ans=min(ans,dp[n][i]);
    cout<<ans<<endl;
}
### 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 <iostream>
using namespace std;
typedef long long LL;
const int N=5022;
LL dp[N][N];
int n,a,b,x[N],pos[N];
int main(){
    scanf("%d%d%d",&n,&a,&b);
    for(int i=1;i<=n;i++) scanf("%d",&x[i]), pos[x[i]]=i;
    for(int i=0;i<N;i++)for(int j=0;j<N;j++)dp[i][j]=1e18;
    dp[0][0]=0;
    for(int i=1;i<=n;i++){
        LL mn=dp[i-1][0];
        for(int j=0;j<=n+1;j++){
            mn=min(mn,dp[i-1][j]);

            if(j==pos[i]) dp[i][j]=mn;
            else if(j<pos[i]) dp[i][j]=mn+min(1LL*b,1LL*a*(pos[i]-j));
            else dp[i][j]=mn+min(1LL*a,1LL*b*(j-pos[i]));
    
            //printf("dp[%d][%d]=%lld\n", i,j,dp[i][j]);
        }
    }
    LL ans=1e18;
    for(int i=0;i<=n+1;i++)ans=min(ans,dp[n][i]);
    cout<<ans<<endl;
}
```