#include<cstdio>
using namespace std;
#define N 5005
long long dp[N][N];
int n,a,b,p[N];
int main()
{
	scanf("%d%d%d",&n,&a,&b);
	for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)dp[i][j]=1e17;
	dp[0][0]=0;
	for(int i=1;i<=n;i++)scanf("%d",&p[i]);
	for(int i=1;i<=n;i++)
	{
		long long mn=1e18;
		for(int j=0;j<=n;j++)
		{
			long long tp=mn;
			if(p[i]>j)tp+=a;
			if(p[i]<j)tp+=b;
			if(dp[i][j]>tp)dp[i][j]=tp;
			if(mn>dp[i-1][j])mn=dp[i-1][j];
			tp=mn;
			if(p[i]>j)tp+=a;else tp+=b;
			if(dp[i][j]>tp)dp[i][j]=tp;
		}
	}
	long long as=1e18;
	for(int i=0;i<=n;i++)if(as>dp[n][i])as=dp[n][i];
	printf("%lld\n",as);
}//
### Prompt

Generate 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>
using namespace std;
#define N 5005
long long dp[N][N];
int n,a,b,p[N];
int main()
{
	scanf("%d%d%d",&n,&a,&b);
	for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)dp[i][j]=1e17;
	dp[0][0]=0;
	for(int i=1;i<=n;i++)scanf("%d",&p[i]);
	for(int i=1;i<=n;i++)
	{
		long long mn=1e18;
		for(int j=0;j<=n;j++)
		{
			long long tp=mn;
			if(p[i]>j)tp+=a;
			if(p[i]<j)tp+=b;
			if(dp[i][j]>tp)dp[i][j]=tp;
			if(mn>dp[i-1][j])mn=dp[i-1][j];
			tp=mn;
			if(p[i]>j)tp+=a;else tp+=b;
			if(dp[i][j]>tp)dp[i][j]=tp;
		}
	}
	long long as=1e18;
	for(int i=0;i<=n;i++)if(as>dp[n][i])as=dp[n][i];
	printf("%lld\n",as);
}//
```