#include <iostream>

using ll = long long;
#define MIN(a, b) (((a) < (b)) ? (a) : (b))
#define MAX(a, b) (((a) < (b)) ? (b) : (a))

int const nmax = 5000;
int const inf = 1000000000;

int v[5 + nmax];
ll dp[5 + nmax];

int main() {
  int n, a, b;
  std::cin >> n >> a >> b;
  for(int i = 1; i <= n; i++)
    std::cin >> v[i];
  v[n + 1] = n + 1;
  for(int i = 1; i <= n + 1; i++) {
    int smax = -1, bigger = 0;
    dp[i] = 1LL * nmax * inf;
    for(int j = i - 1; 0 <= j; j--) {
      if(v[j] < v[i] && smax < v[j]) { 
        dp[i] = std::min(dp[i], dp[j] + 1LL * bigger * a + 1LL * (i - j - 1 - bigger) * b);
      }
      if(v[j] < v[i])
        smax = std::max(smax, v[j]);
      else
        bigger++;
    }
  }
  std::cout << dp[n + 1];
  return 0;
}

### Prompt

Develop a solution in Cpp 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 <iostream>

using ll = long long;
#define MIN(a, b) (((a) < (b)) ? (a) : (b))
#define MAX(a, b) (((a) < (b)) ? (b) : (a))

int const nmax = 5000;
int const inf = 1000000000;

int v[5 + nmax];
ll dp[5 + nmax];

int main() {
  int n, a, b;
  std::cin >> n >> a >> b;
  for(int i = 1; i <= n; i++)
    std::cin >> v[i];
  v[n + 1] = n + 1;
  for(int i = 1; i <= n + 1; i++) {
    int smax = -1, bigger = 0;
    dp[i] = 1LL * nmax * inf;
    for(int j = i - 1; 0 <= j; j--) {
      if(v[j] < v[i] && smax < v[j]) { 
        dp[i] = std::min(dp[i], dp[j] + 1LL * bigger * a + 1LL * (i - j - 1 - bigger) * b);
      }
      if(v[j] < v[i])
        smax = std::max(smax, v[j]);
      else
        bigger++;
    }
  }
  std::cout << dp[n + 1];
  return 0;
}

```