//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
#define N 5010
ll f[N][N]; int a[N];
void upd(ll &x,ll y){x=x<y?x:y;}
int main()
{
	int n=read(),A=read(),B=read();
	for(int i=1;i<=n;i++)	a[i]=read();
	memset(f,48,sizeof(f)); f[0][0]=0;
	for(int i=1;i<=n;i++)	for(int j=0;j<=n;j++)
		if(a[i]>j)	upd(f[i][j],f[i-1][j]+A),upd(f[i][a[i]],f[i-1][j]);
		else	upd(f[i][j],f[i-1][j]+B);
	for(int i=1;i<=n;i++)	upd(f[n][0],f[n][i]);
	printf("%lld\n",f[n][0]);
	return 0;
}

### Prompt

Please provide a CPP coded solution 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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
#define N 5010
ll f[N][N]; int a[N];
void upd(ll &x,ll y){x=x<y?x:y;}
int main()
{
	int n=read(),A=read(),B=read();
	for(int i=1;i<=n;i++)	a[i]=read();
	memset(f,48,sizeof(f)); f[0][0]=0;
	for(int i=1;i<=n;i++)	for(int j=0;j<=n;j++)
		if(a[i]>j)	upd(f[i][j],f[i-1][j]+A),upd(f[i][a[i]],f[i-1][j]);
		else	upd(f[i][j],f[i-1][j]+B);
	for(int i=1;i<=n;i++)	upd(f[n][0],f[n][i]);
	printf("%lld\n",f[n][0]);
	return 0;
}

```