#include<cstdio>
#define N 5005
#define inf 0x3f3f3f3f3f3f3f3fll
inline long long min(long long x,long long y){return x<y?x:y;}
inline long long max(long long x,long long y){return x<y?y:x;}
int a[N],b[N],i,j,n,o,p;long long f[2][N],x;
int main()
{
	scanf("%d%d%d",&n,&o,&p),x=f[1][0]=inf;
	for(i=1;i<=n;i++)scanf("%d",a+i),f[0][b[a[i]]=i]=f[1][i]=inf;
	for(i=1;i<=n;i++)for(j=0;j<=n;f[i&1^1][j++]=inf)
	if(a[i]>j)f[i&1][a[i]]=min(f[i&1][a[i]],f[i&1^1][j]),f[i&1][j]=min(f[i&1][j],f[i&1^1][j]+o);
	else f[i&1][j]=min(f[i&1][j],f[i&1^1][j]+p);
	for(i=0;i<=n;i++)x=min(x,f[n&1][i]);
	return 0*printf("%lld\n",x);
}
### Prompt

Your task is to create 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>
#define N 5005
#define inf 0x3f3f3f3f3f3f3f3fll
inline long long min(long long x,long long y){return x<y?x:y;}
inline long long max(long long x,long long y){return x<y?y:x;}
int a[N],b[N],i,j,n,o,p;long long f[2][N],x;
int main()
{
	scanf("%d%d%d",&n,&o,&p),x=f[1][0]=inf;
	for(i=1;i<=n;i++)scanf("%d",a+i),f[0][b[a[i]]=i]=f[1][i]=inf;
	for(i=1;i<=n;i++)for(j=0;j<=n;f[i&1^1][j++]=inf)
	if(a[i]>j)f[i&1][a[i]]=min(f[i&1][a[i]],f[i&1^1][j]),f[i&1][j]=min(f[i&1][j],f[i&1^1][j]+o);
	else f[i&1][j]=min(f[i&1][j],f[i&1^1][j]+p);
	for(i=0;i<=n;i++)x=min(x,f[n&1][i]);
	return 0*printf("%lld\n",x);
}
```