#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5050;
int n, a, b;
int loc[N];
ll f[N][2 * N];
int main() {
	cin >> n >> a >> b;
	for(int i = 1; i <= n; i++) {
		int t; scanf("%d", &t); loc[t] = i;
	}
	memset(f, 127, sizeof f);
	memset(f[0], 0, sizeof f[0]);
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= 2 * n; j++) {
			f[i][j] = f[i - 1][j - (j & 1)];
			if (j < loc[i] * 2) f[i][j] += b;
				else if (j > loc[i] * 2) f[i][j] += a;
			f[i][j] = min(f[i][j - 1], f[i][j]);
		}
	}
	ll ans = 1e18;
	for(int j = 1; j <= 2 * n; j++) ans = min(ans, f[n][j]);
	cout << ans << endl;
}
### Prompt

Your challenge is to write a Cpp solution to 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5050;
int n, a, b;
int loc[N];
ll f[N][2 * N];
int main() {
	cin >> n >> a >> b;
	for(int i = 1; i <= n; i++) {
		int t; scanf("%d", &t); loc[t] = i;
	}
	memset(f, 127, sizeof f);
	memset(f[0], 0, sizeof f[0]);
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= 2 * n; j++) {
			f[i][j] = f[i - 1][j - (j & 1)];
			if (j < loc[i] * 2) f[i][j] += b;
				else if (j > loc[i] * 2) f[i][j] += a;
			f[i][j] = min(f[i][j - 1], f[i][j]);
		}
	}
	ll ans = 1e18;
	for(int j = 1; j <= 2 * n; j++) ans = min(ans, f[n][j]);
	cout << ans << endl;
}
```