#include <bits/stdc++.h>
using namespace std;

#define LL long long

const int maxN = 5005;

int n;
int cnt[maxN + 1], a[maxN + 1], pos[maxN + 1];
LL A, B;
LL f[maxN + 1][maxN + 1];

int main()
{
	scanf("%d", &n);
	scanf("%lld %lld", &A, &B);
	for(int i = 1; i <= n; i++) 
	{
		scanf("%d", &a[i]);
		pos[ a[i] ] = i;
	}
	for(int i = 1; i <= n; i++)
	{
		for(int j = a[i] + 1; j <= n; j++)
			if(pos[j] <= i) cnt[j] ++;
		for(int j = 1; j <= n; j++)
			if(pos[j] <= i)
			{
				if(!cnt[j]) f[i][j] = f[i][j - 1];
				else f[i][j] = min(f[i][j - 1] + A, f[ pos[j] - 1 ][j] + B * cnt[j]);
			}
			else f[i][j] = f[i][j - 1];
	}
	printf("%lld", f[n][n]);
	return 0;
}

### Prompt

Please formulate 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 LL long long

const int maxN = 5005;

int n;
int cnt[maxN + 1], a[maxN + 1], pos[maxN + 1];
LL A, B;
LL f[maxN + 1][maxN + 1];

int main()
{
	scanf("%d", &n);
	scanf("%lld %lld", &A, &B);
	for(int i = 1; i <= n; i++) 
	{
		scanf("%d", &a[i]);
		pos[ a[i] ] = i;
	}
	for(int i = 1; i <= n; i++)
	{
		for(int j = a[i] + 1; j <= n; j++)
			if(pos[j] <= i) cnt[j] ++;
		for(int j = 1; j <= n; j++)
			if(pos[j] <= i)
			{
				if(!cnt[j]) f[i][j] = f[i][j - 1];
				else f[i][j] = min(f[i][j - 1] + A, f[ pos[j] - 1 ][j] + B * cnt[j]);
			}
			else f[i][j] = f[i][j - 1];
	}
	printf("%lld", f[n][n]);
	return 0;
}

```