#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e3 + 10;

int n, A, B, p[N];
ll dp[N];

int main() {
  scanf("%d %d %d", &n, &A, &B);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  p[++n] = n;
  for (int i = 1; i <= n; i++) {
    dp[i] = (ll) 1e18;
    ll tmp = 0;
    for (int j = i - 1; j >= 0; j--) {
      if (p[j] < p[i]) {
        dp[i] = min(dp[i], dp[j] + tmp);
        tmp += B;
      } else {
        tmp += A;
      }
    }
  }
  printf("%lld\n", dp[n]);
  return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 5e3 + 10;

int n, A, B, p[N];
ll dp[N];

int main() {
  scanf("%d %d %d", &n, &A, &B);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  p[++n] = n;
  for (int i = 1; i <= n; i++) {
    dp[i] = (ll) 1e18;
    ll tmp = 0;
    for (int j = i - 1; j >= 0; j--) {
      if (p[j] < p[i]) {
        dp[i] = min(dp[i], dp[j] + tmp);
        tmp += B;
      } else {
        tmp += A;
      }
    }
  }
  printf("%lld\n", dp[n]);
  return 0;
}
```