#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
using ll = long long;

int main() {
  ll N, A, B;
  cin >> N >> A >> B;
  vector<ll> P;
  P.push_back(0);
  for(int i = 0; i < N; i++) {
    ll p;
    cin >> p;
    P.push_back(p);
  }
  P.push_back(N + 1);
  N = P.size();
  vector<ll> dp(N, 1LL << 62);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    ll acc = 0;
    for(int j = i - 1; j >= 0; j--) {
      if (P[j] < P[i]) {
        dp[i] = min(dp[i], dp[j] + acc);
        acc += B;
      } else {
        acc += A;
      }
    }
  }
  cout << dp.back() << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#include <vector>
#include <algorithm>

using namespace std;
using ll = long long;

int main() {
  ll N, A, B;
  cin >> N >> A >> B;
  vector<ll> P;
  P.push_back(0);
  for(int i = 0; i < N; i++) {
    ll p;
    cin >> p;
    P.push_back(p);
  }
  P.push_back(N + 1);
  N = P.size();
  vector<ll> dp(N, 1LL << 62);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    ll acc = 0;
    for(int j = i - 1; j >= 0; j--) {
      if (P[j] < P[i]) {
        dp[i] = min(dp[i], dp[j] + acc);
        acc += B;
      } else {
        acc += A;
      }
    }
  }
  cout << dp.back() << endl;
  return 0;
}

```