#include<bits/stdc++.h>

using namespace std;

int main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL);

  int n, X, Y; cin >> n >> X >> Y;
  vector<int> a(n);
  for(int &x : a) cin >> x;

  vector<vector<long long>> f(n + 1, vector<long long>(n + 2, 1e18));

  f[0][0] = 0;
  for(int i = 0; i < n; ++i) {
    for(int j = 0; j <= n; ++j) {
      if(j > 0) f[i][j] = min(f[i][j], f[i][j - 1]);

      if(a[i] > j) {
        f[i + 1][a[i]] = min(f[i + 1][a[i]], f[i][j]);
        f[i + 1][j] = min(f[i + 1][j], f[i][j] + X);
      }
      else {
        f[i + 1][j] = min(f[i + 1][j], f[i][j] + Y);
      }
    }
  }

  cout << *min_element(f[n].begin(), f[n].end()) << 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;

int main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL);

  int n, X, Y; cin >> n >> X >> Y;
  vector<int> a(n);
  for(int &x : a) cin >> x;

  vector<vector<long long>> f(n + 1, vector<long long>(n + 2, 1e18));

  f[0][0] = 0;
  for(int i = 0; i < n; ++i) {
    for(int j = 0; j <= n; ++j) {
      if(j > 0) f[i][j] = min(f[i][j], f[i][j - 1]);

      if(a[i] > j) {
        f[i + 1][a[i]] = min(f[i + 1][a[i]], f[i][j]);
        f[i + 1][j] = min(f[i + 1][j], f[i][j] + X);
      }
      else {
        f[i + 1][j] = min(f[i + 1][j], f[i][j] + Y);
      }
    }
  }

  cout << *min_element(f[n].begin(), f[n].end()) << endl;

  return 0;
}

```