#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
 
using namespace std;
 
typedef long long ll;
 
const ll INF = 1e9, MOD = 1e9 + 7;

ll n, d[5001][5001], A, B;

int main ()
{
	cin >> n >> A >> B;

	for (ll i = 1; i <= n; i++)
	{
		ll a;

		scanf ("%lld", &a);

		for (ll j = 0; j <= n; j++)
		{
			if (j < a) d[i][j] = d[i-1][j] + A;
			else if (j == a) d[i][j] = d[i-1][j];
			else d[i][j] = d[i-1][j] + B;
			if (j) d[i][j] = min (d[i][j], d[i][j-1]);
		}
	}

	cout << d[n][n];
}
### Prompt

Generate 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 <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
 
using namespace std;
 
typedef long long ll;
 
const ll INF = 1e9, MOD = 1e9 + 7;

ll n, d[5001][5001], A, B;

int main ()
{
	cin >> n >> A >> B;

	for (ll i = 1; i <= n; i++)
	{
		ll a;

		scanf ("%lld", &a);

		for (ll j = 0; j <= n; j++)
		{
			if (j < a) d[i][j] = d[i-1][j] + A;
			else if (j == a) d[i][j] = d[i-1][j];
			else d[i][j] = d[i-1][j] + B;
			if (j) d[i][j] = min (d[i][j], d[i][j-1]);
		}
	}

	cout << d[n][n];
}
```