#include <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 5005;
int dp[N][N], p[N], cnt[N][N], a, b, n;

int solve(int i, int j) {
  if (i == 0 || j == 0) {
    return 0;
  }
  if (~dp[i][j]) {
    return dp[i][j];
  }
  if (p[i - 1] >= j) {
    return solve(i - 1, j);
  }
  int cur = cnt[i][j] - cnt[i][p[i - 1] + 1];
  return dp[i][j] = min(solve(i - 1, j) + b, solve(i - 1, min(p[i - 1], j)) + cur * a);
}

signed main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> a >> b;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < N; j++) {
      cnt[i + 1][j] = cnt[i][j] + (p[i] < j);
    }
  }
  cout << solve(n, n) << '\n';
  return 0;
}


### Prompt

Generate a CPP solution 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 <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 5005;
int dp[N][N], p[N], cnt[N][N], a, b, n;

int solve(int i, int j) {
  if (i == 0 || j == 0) {
    return 0;
  }
  if (~dp[i][j]) {
    return dp[i][j];
  }
  if (p[i - 1] >= j) {
    return solve(i - 1, j);
  }
  int cur = cnt[i][j] - cnt[i][p[i - 1] + 1];
  return dp[i][j] = min(solve(i - 1, j) + b, solve(i - 1, min(p[i - 1], j)) + cur * a);
}

signed main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> a >> b;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < N; j++) {
      cnt[i + 1][j] = cnt[i][j] + (p[i] < j);
    }
  }
  cout << solve(n, n) << '\n';
  return 0;
}


```