#include <iostream>
using namespace std;
int main() {
    int n, A, B; cin >> n >> A >> B;
    int p[n]; for (int i = 0; i < n; i++) cin >> p[i];
    long long dp[2][n+1];
    int cur = 0, prev = 1;
    fill(dp[prev], dp[prev]+n+1, 0);
    for (int i = n-1; i >= 0; i--) {
        for (int j = 0; j <= n; j++) {
            if (j < p[i]) dp[cur][j] = min(dp[prev][p[i]], dp[prev][j] + A);
            else if (j > p[i]) dp[cur][j] = dp[prev][j] + B;
        }
        swap(cur, prev);
    }
    cout << dp[prev][0] << endl;
}

### 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 <iostream>
using namespace std;
int main() {
    int n, A, B; cin >> n >> A >> B;
    int p[n]; for (int i = 0; i < n; i++) cin >> p[i];
    long long dp[2][n+1];
    int cur = 0, prev = 1;
    fill(dp[prev], dp[prev]+n+1, 0);
    for (int i = n-1; i >= 0; i--) {
        for (int j = 0; j <= n; j++) {
            if (j < p[i]) dp[cur][j] = min(dp[prev][p[i]], dp[prev][j] + A);
            else if (j > p[i]) dp[cur][j] = dp[prev][j] + B;
        }
        swap(cur, prev);
    }
    cout << dp[prev][0] << endl;
}

```