#include <stdio.h>

int N, A, B; long long D[5050];

int main()
{
	scanf ("%d %d %d", &N, &A, &B);
	for (int i = 1; i <= N; i++){
		int x; scanf ("%d", &x);
		long long X[5050];
		for (int j = 0; j <= N; j++) X[j] = 1e18;
		for (int j = 0; j <= N; j++){
			if (j <= x){
				if (X[x] > D[j]) X[x] = D[j];
				if (X[j] > D[j] + A) X[j] = D[j] + A;
			}
			else{
				if (X[j] > D[j] + B) X[j] = D[j] + B;
			}
		}
		for (int j = 0; j <= N; j++) D[j] = X[j];
	}

	long long ans = 1e18;
	for (int j = 0; j <= N; j++) if (ans > D[j]) ans = D[j];
	printf ("%lld\n", ans);

	return 0;
}
### 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 <stdio.h>

int N, A, B; long long D[5050];

int main()
{
	scanf ("%d %d %d", &N, &A, &B);
	for (int i = 1; i <= N; i++){
		int x; scanf ("%d", &x);
		long long X[5050];
		for (int j = 0; j <= N; j++) X[j] = 1e18;
		for (int j = 0; j <= N; j++){
			if (j <= x){
				if (X[x] > D[j]) X[x] = D[j];
				if (X[j] > D[j] + A) X[j] = D[j] + A;
			}
			else{
				if (X[j] > D[j] + B) X[j] = D[j] + B;
			}
		}
		for (int j = 0; j <= N; j++) D[j] = X[j];
	}

	long long ans = 1e18;
	for (int j = 0; j <= N; j++) if (ans > D[j]) ans = D[j];
	printf ("%lld\n", ans);

	return 0;
}
```