#include<bits/stdc++.h>

using namespace std;

const int N = 5678;
const long long llinf = 1e18;

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

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  ++n;
  p[n] = n;
  fill(dp + 1, dp + n + 1, llinf);
  for (int i = 0; i < n; ++i) {
    int foo = n + 1;
    long long cost = 0;
    for (int j = i + 1; j <= n; ++j) {
      if (p[j] > p[i]) {
        if (p[j] < foo) {
          dp[j] = min(dp[j], dp[i] + cost);
        }
        foo = min(foo, p[j]);
      }
      if (p[j] < p[i]) {
        cost += b;
      } else {
        cost += a;
      }
    }
  }
  cout << dp[n] << '\n';
  return 0;
}
### Prompt

Please create a solution in Cpp 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>

using namespace std;

const int N = 5678;
const long long llinf = 1e18;

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

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  ++n;
  p[n] = n;
  fill(dp + 1, dp + n + 1, llinf);
  for (int i = 0; i < n; ++i) {
    int foo = n + 1;
    long long cost = 0;
    for (int j = i + 1; j <= n; ++j) {
      if (p[j] > p[i]) {
        if (p[j] < foo) {
          dp[j] = min(dp[j], dp[i] + cost);
        }
        foo = min(foo, p[j]);
      }
      if (p[j] < p[i]) {
        cost += b;
      } else {
        cost += a;
      }
    }
  }
  cout << dp[n] << '\n';
  return 0;
}
```