#include<iostream>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;
int n,A,B;
LL ans;
const int N=5010;
LL f[N],p[N];
int main()
{
	cin>>n>>A>>B;memset(f,0x3f,sizeof(f));ans=f[0];f[0]=0;
	for(int i=1,x;i<=n;++i)
	{
		scanf("%d",&x);
		for(int j=0;j<=n;++j)
			if(!j)p[j]=f[j];
			else p[j]=min(p[j-1],f[j]);
		for(int j=0;j<=x;++j)f[j]+=A;
		for(int j=x+1;j<=n;++j)f[j]+=B;
		f[x]=min(f[x],p[x]);
	}
	for(int i=0;i<=n;++i)ans=min(ans,f[i]);
	cout<<ans;
	return 0;
}
### Prompt

Please create a solution in cpp 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>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;
int n,A,B;
LL ans;
const int N=5010;
LL f[N],p[N];
int main()
{
	cin>>n>>A>>B;memset(f,0x3f,sizeof(f));ans=f[0];f[0]=0;
	for(int i=1,x;i<=n;++i)
	{
		scanf("%d",&x);
		for(int j=0;j<=n;++j)
			if(!j)p[j]=f[j];
			else p[j]=min(p[j-1],f[j]);
		for(int j=0;j<=x;++j)f[j]+=A;
		for(int j=x+1;j<=n;++j)f[j]+=B;
		f[x]=min(f[x],p[x]);
	}
	for(int i=0;i<=n;++i)ans=min(ans,f[i]);
	cout<<ans;
	return 0;
}
```