#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ll,ii>
#define endl '\n'

int n,a,b;
int arr[5005];
int pos[5005];

ll memo[5005][10005];

int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	
	memset(memo,127,sizeof(memo));
	
	cin>>n>>a>>b;
	for (int x=1;x<=n;x++) cin>>arr[x];
	for (int x=1;x<=n;x++) pos[arr[x]]=x*2;

	for (int x=0;x<=2*n;x++) memo[0][x]=0;
	
	for (int x=1;x<=n;x++){
		for (int y=1;y<=2*n+1;y++){
			if (y%2==0 && pos[x]==y){
				memo[x][y]=min(memo[x][y-1],memo[x-1][y]);
			}
			else{
				memo[x][y]=min(memo[x][y-1],memo[x-1][y]+(pos[x]<y?a:b));
			}
			//cout<<x<<" "<<y<<" "<<memo[x][y]<<endl;
		}
	}
	
	cout<<memo[n][2*n+1];
}
### Prompt

In CPP, your task is to solve 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 ll long long
#define ii pair<ll,ll>
#define iii pair<ll,ii>
#define endl '\n'

int n,a,b;
int arr[5005];
int pos[5005];

ll memo[5005][10005];

int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	
	memset(memo,127,sizeof(memo));
	
	cin>>n>>a>>b;
	for (int x=1;x<=n;x++) cin>>arr[x];
	for (int x=1;x<=n;x++) pos[arr[x]]=x*2;

	for (int x=0;x<=2*n;x++) memo[0][x]=0;
	
	for (int x=1;x<=n;x++){
		for (int y=1;y<=2*n+1;y++){
			if (y%2==0 && pos[x]==y){
				memo[x][y]=min(memo[x][y-1],memo[x-1][y]);
			}
			else{
				memo[x][y]=min(memo[x][y-1],memo[x-1][y]+(pos[x]<y?a:b));
			}
			//cout<<x<<" "<<y<<" "<<memo[x][y]<<endl;
		}
	}
	
	cout<<memo[n][2*n+1];
}
```