#include <bits/stdc++.h>
using namespace std;long long n,aa,i,j,b,f[5005],pos[5005],ans,a[5005],c[5005];int main(){ans=12345678912345678ll;cin>>n>>aa>>b;for (i=1;i<=n;i++) f[i]=ans;for (i=1;i<=n;i++){	cin>>a[i];pos[a[i]]=i;	}	for (i=1;i<=n;i++){		c[i]=c[i-1]+(i<a[i]?aa:b);	}	for (i=1;i<=n;i++){		long long s=0;		for (j=i-1;j>=0;j--){			if (a[j]<a[i]){				f[i]=min(f[i],f[j]+s);			}			if (a[j]<a[i]){				s+=b;			} else {				s+=aa;			}		}		ans=min(ans,f[i]+(n-i)*b);	}	cout<<ans<<endl;return 0;}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;long long n,aa,i,j,b,f[5005],pos[5005],ans,a[5005],c[5005];int main(){ans=12345678912345678ll;cin>>n>>aa>>b;for (i=1;i<=n;i++) f[i]=ans;for (i=1;i<=n;i++){	cin>>a[i];pos[a[i]]=i;	}	for (i=1;i<=n;i++){		c[i]=c[i-1]+(i<a[i]?aa:b);	}	for (i=1;i<=n;i++){		long long s=0;		for (j=i-1;j>=0;j--){			if (a[j]<a[i]){				f[i]=min(f[i],f[j]+s);			}			if (a[j]<a[i]){				s+=b;			} else {				s+=aa;			}		}		ans=min(ans,f[i]+(n-i)*b);	}	cout<<ans<<endl;return 0;}
```