#include<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
ll f[5010][5010],n,A,B,p[5010];
int main(){
	read(n);read(A);read(B);
	for(int i=1;i<=n;i++) read(p[i]);
	memset(f,63,sizeof(f));
	for(int i=1;i<=n;i++){
		f[i][p[i]]=(i-1)*A;
		for(int j=1;j<=n;j++){
			f[i][j]=min(f[i][j],f[i-1][j]+(p[i]>j?A:B));
			if(j<p[i]) f[i][p[i]]=min(f[i][p[i]],f[i-1][j]);
		}
	}
	ll ans=1e18;
	for(int i=1;i<=n;i++) ans=min(ans,f[n][i]);
	cout<<ans<<'\n';
	return 0;
}

### 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<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
ll f[5010][5010],n,A,B,p[5010];
int main(){
	read(n);read(A);read(B);
	for(int i=1;i<=n;i++) read(p[i]);
	memset(f,63,sizeof(f));
	for(int i=1;i<=n;i++){
		f[i][p[i]]=(i-1)*A;
		for(int j=1;j<=n;j++){
			f[i][j]=min(f[i][j],f[i-1][j]+(p[i]>j?A:B));
			if(j<p[i]) f[i][p[i]]=min(f[i][p[i]],f[i-1][j]);
		}
	}
	ll ans=1e18;
	for(int i=1;i<=n;i++) ans=min(ans,f[n][i]);
	cout<<ans<<'\n';
	return 0;
}

```