#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
int n,v[5005];
ll a,b,f[5005];
int main()
{
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i=1;i<=n+1;i++)
	{
		f[i]=1000000000000000000;
		if(i<=n)scanf("%d",&v[i]);
		else v[i]=n+1;
		ll sum=0;
		int maxx=0;
		for(int j=i-1;j>=1;j--)
		{
			if(v[j]>v[i])sum+=a;
			else if(v[j]>maxx)
			{
				f[i]=min(f[i],f[j]+sum);
				sum+=b;
			}
			else sum+=b;
		}
		f[i]=min(f[i],sum);
	}
	printf("%lld\n",f[n+1]);
	return 0;
}
### Prompt

Please formulate 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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
int n,v[5005];
ll a,b,f[5005];
int main()
{
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i=1;i<=n+1;i++)
	{
		f[i]=1000000000000000000;
		if(i<=n)scanf("%d",&v[i]);
		else v[i]=n+1;
		ll sum=0;
		int maxx=0;
		for(int j=i-1;j>=1;j--)
		{
			if(v[j]>v[i])sum+=a;
			else if(v[j]>maxx)
			{
				f[i]=min(f[i],f[j]+sum);
				sum+=b;
			}
			else sum+=b;
		}
		f[i]=min(f[i],sum);
	}
	printf("%lld\n",f[n+1]);
	return 0;
}
```