#include <bits/stdc++.h>
using namespace std;
int n,a,b,p[5005],idx[5005][5005],cum[5005][5005];
long long dp[5005][5005];
long long solve(int i,int j)
{
	if (!idx[i][j])
	return 0;
	if (dp[i][j]!=-1)
	return dp[i][j];
	return dp[i][j]=min(a+solve(i,p[idx[i][j]]-1),1LL*b*(cum[i][j]-cum[idx[i][j]][j])+solve(idx[i][j]-1,j));
}
int main()
{
	scanf("%d%d%d",&n,&a,&b);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&p[i]);
		for (int j=1;j<=n;j++)
		{
			idx[i][j]=idx[i-1][j];
			if (p[i]<=j && p[i]>p[idx[i][j]])
			idx[i][j]=i;
			cum[i][j]=cum[i-1][j]+(p[i]<=j);
		}
	}
	memset(dp,-1,sizeof(dp));
	printf("%lld",solve(n,n));
}
### Prompt

Develop a solution in Cpp 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;
int n,a,b,p[5005],idx[5005][5005],cum[5005][5005];
long long dp[5005][5005];
long long solve(int i,int j)
{
	if (!idx[i][j])
	return 0;
	if (dp[i][j]!=-1)
	return dp[i][j];
	return dp[i][j]=min(a+solve(i,p[idx[i][j]]-1),1LL*b*(cum[i][j]-cum[idx[i][j]][j])+solve(idx[i][j]-1,j));
}
int main()
{
	scanf("%d%d%d",&n,&a,&b);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&p[i]);
		for (int j=1;j<=n;j++)
		{
			idx[i][j]=idx[i-1][j];
			if (p[i]<=j && p[i]>p[idx[i][j]])
			idx[i][j]=i;
			cum[i][j]=cum[i-1][j]+(p[i]<=j);
		}
	}
	memset(dp,-1,sizeof(dp));
	printf("%lld",solve(n,n));
}
```