#include <bits/stdc++.h>

using namespace std;

#define int long long

int n,a,b;
int p[5001], f[10011], mn[10011], pos[5001];

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> a >> b;
	for (int i=1;i<=n;i++)
	{
		cin >> p[i];
		pos[p[i]] = i*2;
	}
	mn[0] = 1e18;
	for (int j=1;j<=n;j++)
	{
    	for (int i=1;i<=2*n+1;i++)
    	{
    		int cost = 0;
    		if (i < pos[j]) cost = b;
    		else if (i > pos[j]) cost = a;
    		if (i%2 == 1)
    			f[i] = mn[i] + cost;			
    		else
    			f[i] = mn[i-1] + cost;
    	}
		for (int i=1;i<=2*n+1;i++)
			mn[i] = min(mn[i-1],f[i]);
	}
	cout << mn[2*n+1];	
}
### Prompt

Develop a solution in Cpp 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;

#define int long long

int n,a,b;
int p[5001], f[10011], mn[10011], pos[5001];

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> a >> b;
	for (int i=1;i<=n;i++)
	{
		cin >> p[i];
		pos[p[i]] = i*2;
	}
	mn[0] = 1e18;
	for (int j=1;j<=n;j++)
	{
    	for (int i=1;i<=2*n+1;i++)
    	{
    		int cost = 0;
    		if (i < pos[j]) cost = b;
    		else if (i > pos[j]) cost = a;
    		if (i%2 == 1)
    			f[i] = mn[i] + cost;			
    		else
    			f[i] = mn[i-1] + cost;
    	}
		for (int i=1;i<=2*n+1;i++)
			mn[i] = min(mn[i-1],f[i]);
	}
	cout << mn[2*n+1];	
}
```