#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0;i<((int)(n));i++)
#define reg(i,a,b) for(int i = ((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i = ((int)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(int i = ((int)(b));i>=((int)(a));i--)
typedef long long ll;
typedef pair<ll, ll> mp;


/*
1~nを左から順に配置していく
dp(i,j)=iをjに置く
*/

ll n,A,B,a[5010],dp[5010]={},inf=1e18,ans=1e18;

int main(void){
	cin>>n>>A>>B;
	reg(i,1,n){
		ll x;
		cin>>x;
		a[x]=i;
	}
	reg(i,1,n){
		ll c=inf;
		reg(j,1,n){
			c=min(c,dp[j]);
			if(a[i]<j){
				dp[j]=c+A;
			}else if(a[i]>j){
				dp[j]=c+B;
			}else{
				dp[j]=c;
			}
		}
	}
	reg(i,1,n)ans=min(ans,dp[i]);
	cout<<ans<<endl;
	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>
using namespace std;
#define rep(i,n) for(int i = 0;i<((int)(n));i++)
#define reg(i,a,b) for(int i = ((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i = ((int)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(int i = ((int)(b));i>=((int)(a));i--)
typedef long long ll;
typedef pair<ll, ll> mp;


/*
1~nを左から順に配置していく
dp(i,j)=iをjに置く
*/

ll n,A,B,a[5010],dp[5010]={},inf=1e18,ans=1e18;

int main(void){
	cin>>n>>A>>B;
	reg(i,1,n){
		ll x;
		cin>>x;
		a[x]=i;
	}
	reg(i,1,n){
		ll c=inf;
		reg(j,1,n){
			c=min(c,dp[j]);
			if(a[i]<j){
				dp[j]=c+A;
			}else if(a[i]>j){
				dp[j]=c+B;
			}else{
				dp[j]=c;
			}
		}
	}
	reg(i,1,n)ans=min(ans,dp[i]);
	cout<<ans<<endl;
	return 0;
}
```