#include <bits/stdc++.h>

using namespace std;

const int N = 5050;

int a[N];
long long f[N];

int main() {
	int n, A, B;
	scanf("%d %d %d", &n, &A, &B);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	a[0] = 0;
	a[n + 1] = n + 1;
	f[0] = 0;
	for (int i = 1; i <= n + 1; ++i) {
		f[i] = 1e18;
		int cnt = 0;
		for (int j = i - 1; j >= 0; --j) {
			if (a[i] > a[j]) {
				f[i] = min(f[i], f[j] + (long long) A * cnt + (long long) B * (i - j - 1 - cnt));
			} else {
				++cnt;
			}
		}
	}
	printf("%lld\n", f[n + 1]);	
}

### 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;

const int N = 5050;

int a[N];
long long f[N];

int main() {
	int n, A, B;
	scanf("%d %d %d", &n, &A, &B);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	a[0] = 0;
	a[n + 1] = n + 1;
	f[0] = 0;
	for (int i = 1; i <= n + 1; ++i) {
		f[i] = 1e18;
		int cnt = 0;
		for (int j = i - 1; j >= 0; --j) {
			if (a[i] > a[j]) {
				f[i] = min(f[i], f[j] + (long long) A * cnt + (long long) B * (i - j - 1 - cnt));
			} else {
				++cnt;
			}
		}
	}
	printf("%lld\n", f[n + 1]);	
}

```