#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long 
ll dp[5050],A,B;
int a[5050],n;
int main(){
	memset(dp,63,sizeof dp);
	scanf("%d",&n);
	scanf("%lld%lld",&A,&B);
	for (int i=1;i<=n;i++)scanf("%d",&a[i]);
	a[n+1]=n+1;
	dp[0]=0;
	for (int i=1;i<=n+1;i++){
		int g=0;
		for (int j=i-1;j>=0;j--)
			if (a[j]>a[i])g++;
			else dp[i]=std::min(dp[i],dp[j]+A*g+B*(i-j-1-g));
	}
	printf("%lld",dp[n+1]);
} 
### 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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long 
ll dp[5050],A,B;
int a[5050],n;
int main(){
	memset(dp,63,sizeof dp);
	scanf("%d",&n);
	scanf("%lld%lld",&A,&B);
	for (int i=1;i<=n;i++)scanf("%d",&a[i]);
	a[n+1]=n+1;
	dp[0]=0;
	for (int i=1;i<=n+1;i++){
		int g=0;
		for (int j=i-1;j>=0;j--)
			if (a[j]>a[i])g++;
			else dp[i]=std::min(dp[i],dp[j]+A*g+B*(i-j-1-g));
	}
	printf("%lld",dp[n+1]);
} 
```