#include <bits/stdc++.h>
const int N=5005;
long long dp[N][N],ans=100000000000000;
int a[N],x,A,B,n; 
int main(){
	scanf("%d%d%d",&n,&A,&B);
	for (int i=1;i<=n;i++){
		scanf("%d",&x);
		a[x]=i;
	}
	memset(dp,0x3f,sizeof(dp));
	for (int i=0;i<=n;i++) dp[0][i]=0;
	for (int i=1;i<=n;i++)
		for (int j=0;j<i;j++){
			if (a[i]<a[j])
				dp[i][a[j]]=std::min(dp[i][a[j]],dp[i-1][a[j]]+A);
			else dp[i][a[j]]=std::min(dp[i][a[j]],dp[i-1][a[j]]+B),
			dp[i][a[i]]=std::min(dp[i][a[i]],dp[i-1][a[j]]);
			
		}
	for (int i=1;i<=n;i++) ans=std::min(ans,dp[n][i]);
	printf("%lld\n",ans); 
} 
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
const int N=5005;
long long dp[N][N],ans=100000000000000;
int a[N],x,A,B,n; 
int main(){
	scanf("%d%d%d",&n,&A,&B);
	for (int i=1;i<=n;i++){
		scanf("%d",&x);
		a[x]=i;
	}
	memset(dp,0x3f,sizeof(dp));
	for (int i=0;i<=n;i++) dp[0][i]=0;
	for (int i=1;i<=n;i++)
		for (int j=0;j<i;j++){
			if (a[i]<a[j])
				dp[i][a[j]]=std::min(dp[i][a[j]],dp[i-1][a[j]]+A);
			else dp[i][a[j]]=std::min(dp[i][a[j]],dp[i-1][a[j]]+B),
			dp[i][a[i]]=std::min(dp[i][a[i]],dp[i-1][a[j]]);
			
		}
	for (int i=1;i<=n;i++) ans=std::min(ans,dp[n][i]);
	printf("%lld\n",ans); 
} 
```