#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[5010][5010], _min[5010][5010];
int n, A, B;
int a[5010], pos[5010];
int main() {
    cin >> n >> A >> B;
    for(int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= n; j++) {
            if(j != pos[i]) {
                dp[i][j] = _min[i - 1][j] + (j > pos[i] ? A : B);
            }
            else if(j == pos[i]) {
                dp[i][j] = min(_min[i - 1][j] + B, _min[i - 1][j - 1]);
            }
            if(j == 0) _min[i][j] = dp[i][j];
            else _min[i][j] = min(_min[i][j - 1], dp[i][j]);
        }
    }
    return cout << _min[n][n] << endl, 0;
}
### Prompt

Please create a solution in cpp 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;
ll dp[5010][5010], _min[5010][5010];
int n, A, B;
int a[5010], pos[5010];
int main() {
    cin >> n >> A >> B;
    for(int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= n; j++) {
            if(j != pos[i]) {
                dp[i][j] = _min[i - 1][j] + (j > pos[i] ? A : B);
            }
            else if(j == pos[i]) {
                dp[i][j] = min(_min[i - 1][j] + B, _min[i - 1][j - 1]);
            }
            if(j == 0) _min[i][j] = dp[i][j];
            else _min[i][j] = min(_min[i][j - 1], dp[i][j]);
        }
    }
    return cout << _min[n][n] << endl, 0;
}
```