#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,A,B;
int a[5005];
long long dp[5005];
int main()
{
	n=read(),A=read(),B=read();
	for(int i=1;i<=n;i++) a[i]=read();
	a[n+1]=0x3f3f3f3f;memset(dp,0x3f,sizeof(dp));dp[0]=0;
	for(int i=1;i<=n+1;i++)
	{
		int sl=0,sr=0;
		for(int j=i-1;j>=0;j--)
		{
			if(a[j]<a[i]) dp[i]=min(dp[i],dp[j]+1ll*sl*A+1ll*sr*B);
			if(a[j]>a[i]) sl++;if(a[j]<a[i]) sr++;
		}
	}
	printf("%lld\n",dp[n+1]);
}
### Prompt

In CPP, your task is to solve 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<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,A,B;
int a[5005];
long long dp[5005];
int main()
{
	n=read(),A=read(),B=read();
	for(int i=1;i<=n;i++) a[i]=read();
	a[n+1]=0x3f3f3f3f;memset(dp,0x3f,sizeof(dp));dp[0]=0;
	for(int i=1;i<=n+1;i++)
	{
		int sl=0,sr=0;
		for(int j=i-1;j>=0;j--)
		{
			if(a[j]<a[i]) dp[i]=min(dp[i],dp[j]+1ll*sl*A+1ll*sr*B);
			if(a[j]>a[i]) sl++;if(a[j]<a[i]) sr++;
		}
	}
	printf("%lld\n",dp[n+1]);
}
```