#include<stdio.h>
int p[8192],wz[8192];
long long dp[5005][5005];
int main()
{
	int n,a,b;
	scanf("%d%d%d",&n,&a,&b);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&p[i]);
		wz[p[i]]=i;
	}
	for(int i=0;i<=n;i++)
	{
		for(int j=0;j<=n;j++)
		{
			if(i==0&&j==0)
			{
				dp[i][j]=0;
				continue;
			}
			dp[i][j]=1000000000000000000LL;
			if(i)
			{
				if(dp[i-1][j]<dp[i][j])
				{
					dp[i][j]=dp[i-1][j];
				}
			}
			if(j)
			{
				int inc=a;
				if(wz[j]>i)
				{
					inc=b;
				}
				if(dp[i][j-1]+inc<dp[i][j])
				{
					dp[i][j]=dp[i][j-1]+inc;
				}
			}
			if(i&&j)
			{
				if(wz[j]==i)
				{
					if(dp[i-1][j-1]<dp[i][j])
					{
						dp[i][j]=dp[i-1][j-1];
					}
				}
			}
		}
	}
	printf("%lld\n",dp[n][n]);
	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<stdio.h>
int p[8192],wz[8192];
long long dp[5005][5005];
int main()
{
	int n,a,b;
	scanf("%d%d%d",&n,&a,&b);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&p[i]);
		wz[p[i]]=i;
	}
	for(int i=0;i<=n;i++)
	{
		for(int j=0;j<=n;j++)
		{
			if(i==0&&j==0)
			{
				dp[i][j]=0;
				continue;
			}
			dp[i][j]=1000000000000000000LL;
			if(i)
			{
				if(dp[i-1][j]<dp[i][j])
				{
					dp[i][j]=dp[i-1][j];
				}
			}
			if(j)
			{
				int inc=a;
				if(wz[j]>i)
				{
					inc=b;
				}
				if(dp[i][j-1]+inc<dp[i][j])
				{
					dp[i][j]=dp[i][j-1]+inc;
				}
			}
			if(i&&j)
			{
				if(wz[j]==i)
				{
					if(dp[i-1][j-1]<dp[i][j])
					{
						dp[i][j]=dp[i-1][j-1];
					}
				}
			}
		}
	}
	printf("%lld\n",dp[n][n]);
	return 0;
}
```