#include<bits/stdc++.h>
#define maxn 5005
using namespace std;
typedef long long LL;
int n,a[maxn];
LL A,B;
LL f[maxn][maxn];
int main()  {
    scanf("%d%lld%lld",&n,&A,&B);
    for (int i=1;i<=n;++i)
        scanf("%d",a+i);
    memset(f,0x3f,sizeof(f));
    f[0][0]=0;
    for (int i=1;i<=n;++i)  {
        for (int j=i-1,c=0;~j;--j)
            if (a[j]<a[i])
                // cout<<i<<" <- "<<j<<endl,
                f[i][i]=min(f[i][i],f[j][j]+(i-j-1-c)*B+c*A);
            else    ++c;
        for (int j=0;j<i;++j)
            f[i][j]=f[i-1][j]+B;
        // for (int j=0;j<=i;++j)
        //     printf("f[%d][%d]=%lld\n",i,j,f[i][j]);
    }
    LL ans=1LL<<62;
    for (int i=1;i<=n;++i)
        ans=min(ans,f[n][i]);
    printf("%lld\n",ans);
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define maxn 5005
using namespace std;
typedef long long LL;
int n,a[maxn];
LL A,B;
LL f[maxn][maxn];
int main()  {
    scanf("%d%lld%lld",&n,&A,&B);
    for (int i=1;i<=n;++i)
        scanf("%d",a+i);
    memset(f,0x3f,sizeof(f));
    f[0][0]=0;
    for (int i=1;i<=n;++i)  {
        for (int j=i-1,c=0;~j;--j)
            if (a[j]<a[i])
                // cout<<i<<" <- "<<j<<endl,
                f[i][i]=min(f[i][i],f[j][j]+(i-j-1-c)*B+c*A);
            else    ++c;
        for (int j=0;j<i;++j)
            f[i][j]=f[i-1][j]+B;
        // for (int j=0;j<=i;++j)
        //     printf("f[%d][%d]=%lld\n",i,j,f[i][j]);
    }
    LL ans=1LL<<62;
    for (int i=1;i<=n;++i)
        ans=min(ans,f[n][i]);
    printf("%lld\n",ans);
    return 0;
}
```