#include<bits/stdc++.h>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=5005;
int n,a[N];
ll f[N],A,B;
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>A>>B;
	fo(i,1,n)cin>>a[i];
	a[n+1]=n+1;
	fo(i,1,n+1){
		int ct=0,ct2=0,mx=0;
		f[i]=1e18;
		fd(j,0,i-1)
			if(a[j]<a[i]){
				if(a[j]<mx){
					++ct2;
					continue;
				}
				f[i]=min(f[i],f[j]+A*ct+B*ct2);
				++ct2;mx=a[j];
			}else ++ct;
	}
	cout<<f[n+1];
}
### 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>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=5005;
int n,a[N];
ll f[N],A,B;
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>A>>B;
	fo(i,1,n)cin>>a[i];
	a[n+1]=n+1;
	fo(i,1,n+1){
		int ct=0,ct2=0,mx=0;
		f[i]=1e18;
		fd(j,0,i-1)
			if(a[j]<a[i]){
				if(a[j]<mx){
					++ct2;
					continue;
				}
				f[i]=min(f[i],f[j]+A*ct+B*ct2);
				++ct2;mx=a[j];
			}else ++ct;
	}
	cout<<f[n+1];
}
```