#include<bits/stdc++.h>
using namespace std;
#define MN 5000
#define ll long long
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
ll f[MN+5],A,B;
int n;
int main()
{
    n=read();A=read();B=read();
    memset(f,63,sizeof(f));f[0]=0;
    for(int i=1;i<=n;++i)
    {
        int x=read();
        for(int j=0;j<x;++j)
        {
            f[x]=min(f[x],f[j]);
            f[j]+=A;
        }
        for(int j=x;++j<=n;) f[j]+=B;
    }
    ll ans=1e18;
    for(int i=0;i<=n;++i) ans=min(ans,f[i]);
    cout<<ans;
    return 0;
}
### Prompt

Your task is to create 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 MN 5000
#define ll long long
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
ll f[MN+5],A,B;
int n;
int main()
{
    n=read();A=read();B=read();
    memset(f,63,sizeof(f));f[0]=0;
    for(int i=1;i<=n;++i)
    {
        int x=read();
        for(int j=0;j<x;++j)
        {
            f[x]=min(f[x],f[j]);
            f[j]+=A;
        }
        for(int j=x;++j<=n;) f[j]+=B;
    }
    ll ans=1e18;
    for(int i=0;i<=n;++i) ans=min(ans,f[i]);
    cout<<ans;
    return 0;
}
```