#include <cstdio>
#include <algorithm>

typedef long long LL;
const int MN = 5005;

int A, B, N, pos[MN];
LL f[MN * 2], g[MN * 2];

int main() {
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1, x; i <= N; ++i)
		scanf("%d", &x), pos[x] = i;
	for (int i = 1; i <= N; ++i) {
		for (int j = 1; j <= N * 2 + 1; ++j)
			g[j] = j & 1 ? j > 1 ? std::min({g[j - 2], f[j - 1], f[j]}) : f[1] : g[j - 1];
		for (int j = 1; j <= N * 2 + 1; ++j)
			g[j] += j > 2 * pos[i] ? A : j < 2 * pos[i] ? B : 0;
		std::swap(f, g);
	}
	printf("%lld\n", *std::min_element(f + 1, f + 2 * N + 2));
	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 <cstdio>
#include <algorithm>

typedef long long LL;
const int MN = 5005;

int A, B, N, pos[MN];
LL f[MN * 2], g[MN * 2];

int main() {
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1, x; i <= N; ++i)
		scanf("%d", &x), pos[x] = i;
	for (int i = 1; i <= N; ++i) {
		for (int j = 1; j <= N * 2 + 1; ++j)
			g[j] = j & 1 ? j > 1 ? std::min({g[j - 2], f[j - 1], f[j]}) : f[1] : g[j - 1];
		for (int j = 1; j <= N * 2 + 1; ++j)
			g[j] += j > 2 * pos[i] ? A : j < 2 * pos[i] ? B : 0;
		std::swap(f, g);
	}
	printf("%lld\n", *std::min_element(f + 1, f + 2 * N + 2));
	return 0;
}
```