#include <bits/stdc++.h>
#define ll long long
#define MAX 5005
using namespace std;

int n, a[MAX];
ll f[MAX], A, B;

int main()
{
    cin >> n >> A >> B;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
    }
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    a[n+1] = n+1;
    n++;
    for(int i = 1; i <= n; i++){
        ll s = 0;
        for(int j = i-1; j >= 0; j--){
            if(a[i] > a[j]){
                f[i] = min(f[i], f[j]+s);
                s += B;
            }
            else s += A;
        }
    }
    cout << f[n] << endl;

	return 0;
}
### Prompt

Your task is to create 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>
#define ll long long
#define MAX 5005
using namespace std;

int n, a[MAX];
ll f[MAX], A, B;

int main()
{
    cin >> n >> A >> B;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
    }
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    a[n+1] = n+1;
    n++;
    for(int i = 1; i <= n; i++){
        ll s = 0;
        for(int j = i-1; j >= 0; j--){
            if(a[i] > a[j]){
                f[i] = min(f[i], f[j]+s);
                s += B;
            }
            else s += A;
        }
    }
    cout << f[n] << endl;

	return 0;
}
```