#include <bits/stdc++.h>
using namespace std;

const int N = 5e3 + 10;

int n, a, b, p[N];
long long f[N][N];

int main(void) {
  scanf("%d%d%d", &n, &a, &b);
  for(int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  memset(f, 0x3f, sizeof f);
  f[0][0] = 0;
  for(int i = 1; i <= n; i++) {
    for(int j = 0, t; j <= n; j++) {
      if(p[i] > j) {
	f[i][p[i]] = min(f[i][p[i]], f[i - 1][j]);
	f[i][j] = min(f[i][j], f[i - 1][j] + a);
      } else {
	f[i][j] = min(f[i][j], f[i - 1][j] + b);
      }
    }
  }
  long long ans = f[0][1];
  for(int i = 0; i <= n; i++) {
    ans = min(ans, f[n][i]);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;

const int N = 5e3 + 10;

int n, a, b, p[N];
long long f[N][N];

int main(void) {
  scanf("%d%d%d", &n, &a, &b);
  for(int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  memset(f, 0x3f, sizeof f);
  f[0][0] = 0;
  for(int i = 1; i <= n; i++) {
    for(int j = 0, t; j <= n; j++) {
      if(p[i] > j) {
	f[i][p[i]] = min(f[i][p[i]], f[i - 1][j]);
	f[i][j] = min(f[i][j], f[i - 1][j] + a);
      } else {
	f[i][j] = min(f[i][j], f[i - 1][j] + b);
      }
    }
  }
  long long ans = f[0][1];
  for(int i = 0; i <= n; i++) {
    ans = min(ans, f[n][i]);
  }
  cout << ans << endl;
  return 0;
}

```