#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5005;
const ll inf=0x3f3f3f3f3f3f3f3fll;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,A,B,pos[N];
ll f[N],ans;
int main(){
	n=read();A=read();B=read();
	for (int i=1;i<=n;i++) pos[read()]=i-1;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=n;j++) f[j]=min(f[j],f[j-1]);
		for (int j=0;j<=n;j++) f[j]+=(pos[i]==j) ? 0 : ((pos[i]<j) ? A : B);
	}
	ans=inf;
	for (int i=0;i<=n;i++) ans=min(ans,f[i]);
	printf("%lld\n",ans);
	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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5005;
const ll inf=0x3f3f3f3f3f3f3f3fll;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,A,B,pos[N];
ll f[N],ans;
int main(){
	n=read();A=read();B=read();
	for (int i=1;i<=n;i++) pos[read()]=i-1;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=n;j++) f[j]=min(f[j],f[j-1]);
		for (int j=0;j<=n;j++) f[j]+=(pos[i]==j) ? 0 : ((pos[i]<j) ? A : B);
	}
	ans=inf;
	for (int i=0;i<=n;i++) ans=min(ans,f[i]);
	printf("%lld\n",ans);
	return 0;
}
```