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

namespace TYC
{
	typedef long long ll;
	const int N = 5005;

	int n, A, B, P[N];
	ll F[N];

	void work()
	{
		scanf("%d%d%d", &n, &A, &B);
		for (int i = 1; i <= n; i++)
			scanf("%d", &P[i]);
		P[n + 1] = n + 1;
		memset(F, 0x3f, sizeof(F));
		F[0] = 0;
		for (int i = 1; i <= n + 1; i++)
		{
			int cnt[2] = {0, 0};
			if (P[i - 1] < P[i])
				F[i] = F[i - 1];
			for (int j = i - 1; j; j--)
			{
				cnt[P[j] < P[i]]++;
				if (P[j - 1] < P[i])
					F[i] = min(F[i], F[j - 1] + (ll)cnt[0] * A + (ll)cnt[1] * B);
			}
		}
		printf("%lld\n", F[n + 1]);
	}
}

int main()
{
	TYC::work();
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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;

namespace TYC
{
	typedef long long ll;
	const int N = 5005;

	int n, A, B, P[N];
	ll F[N];

	void work()
	{
		scanf("%d%d%d", &n, &A, &B);
		for (int i = 1; i <= n; i++)
			scanf("%d", &P[i]);
		P[n + 1] = n + 1;
		memset(F, 0x3f, sizeof(F));
		F[0] = 0;
		for (int i = 1; i <= n + 1; i++)
		{
			int cnt[2] = {0, 0};
			if (P[i - 1] < P[i])
				F[i] = F[i - 1];
			for (int j = i - 1; j; j--)
			{
				cnt[P[j] < P[i]]++;
				if (P[j - 1] < P[i])
					F[i] = min(F[i], F[j - 1] + (ll)cnt[0] * A + (ll)cnt[1] * B);
			}
		}
		printf("%lld\n", F[n + 1]);
	}
}

int main()
{
	TYC::work();
	return 0;
}
```