#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 5003
#define ll long long

using namespace std;

int n, m, a, b;
int pos[N];
ll f[N][N << 1];

int main(){
	cin >> n >> a >> b; m = n * 2 + 1;
	for(int i = 1; i <= n; ++i){
		int x; scanf("%d", &x);
		pos[x] = i * 2;
	}
	memset(f, 127, sizeof f);
	for(int j = 1; j <= m; ++j) f[0][j] = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j){
			ll tmp = (j != pos[i]) * (j < pos[i] ? b : a);
			tmp += f[i - 1][j - !(j & 1)]; 
			f[i][j] = min(f[i][j - 1], tmp);
		}
	cout << f[n][m] << endl;
	return 0;
}
### Prompt

Create a solution in CPP for 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 <cstring>
#include <iostream>
#include <algorithm>
#define N 5003
#define ll long long

using namespace std;

int n, m, a, b;
int pos[N];
ll f[N][N << 1];

int main(){
	cin >> n >> a >> b; m = n * 2 + 1;
	for(int i = 1; i <= n; ++i){
		int x; scanf("%d", &x);
		pos[x] = i * 2;
	}
	memset(f, 127, sizeof f);
	for(int j = 1; j <= m; ++j) f[0][j] = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j){
			ll tmp = (j != pos[i]) * (j < pos[i] ? b : a);
			tmp += f[i - 1][j - !(j & 1)]; 
			f[i][j] = min(f[i][j - 1], tmp);
		}
	cout << f[n][m] << endl;
	return 0;
}
```