#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, w[5010];
long long D[5010], A, B, S[5010];
int main() {
	int i, j;
	scanf("%d%lld%lld", &n, &A, &B);
	for (i = 1; i <= n; i++) {
		scanf("%d", &w[i]);
		D[i] = 1e18;
	}
	n++;
	w[n] = n;
	D[n] = 1e18;
	for (i = 1; i <= n; i++) {
		D[i] = min(D[i], A * (i - 1));
		long long s = 0;
		for (j = i + 1; j <= n; j++) {
			if (w[j] > w[i]) {
				D[j] = min(D[j], D[i] + s);
			}
			if (w[i] < w[j]) {
				s += A;
			}
			else {
				s += B;
			}
		}
	}
	printf("%lld\n", D[n]);
}
### Prompt

In Cpp, your task is to solve 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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, w[5010];
long long D[5010], A, B, S[5010];
int main() {
	int i, j;
	scanf("%d%lld%lld", &n, &A, &B);
	for (i = 1; i <= n; i++) {
		scanf("%d", &w[i]);
		D[i] = 1e18;
	}
	n++;
	w[n] = n;
	D[n] = 1e18;
	for (i = 1; i <= n; i++) {
		D[i] = min(D[i], A * (i - 1));
		long long s = 0;
		for (j = i + 1; j <= n; j++) {
			if (w[j] > w[i]) {
				D[j] = min(D[j], D[i] + s);
			}
			if (w[i] < w[j]) {
				s += A;
			}
			else {
				s += B;
			}
		}
	}
	printf("%lld\n", D[n]);
}
```