#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

signed main() {
#ifdef LOCAL
  assert(freopen("d.in", "r", stdin));
#endif
  int n, a, b;
  cin >> n >> a >> b;
  vector<int> p(n + 1);
  vector<int> pos(n + 1);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
    pos[p[i]] = i;
  }
  p[n] = pos[n] = n;
  vector<ll> d(n + 1);
  for (int i = 0; i < n + 1; ++i) {
    d[i] = 1e18;
    ll cost = 0;
    for (int j = i - 1; j >= 0; --j) {
      if (pos[j] < pos[i]) {
        d[i] = min(d[i], d[j] + cost);
        cost += a;
      } else {
        cost += b;
      }
    }
    d[i] = min(d[i], cost);
  }
  cout << d[n] << '\n';
}

### 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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

signed main() {
#ifdef LOCAL
  assert(freopen("d.in", "r", stdin));
#endif
  int n, a, b;
  cin >> n >> a >> b;
  vector<int> p(n + 1);
  vector<int> pos(n + 1);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
    pos[p[i]] = i;
  }
  p[n] = pos[n] = n;
  vector<ll> d(n + 1);
  for (int i = 0; i < n + 1; ++i) {
    d[i] = 1e18;
    ll cost = 0;
    for (int j = i - 1; j >= 0; --j) {
      if (pos[j] < pos[i]) {
        d[i] = min(d[i], d[j] + cost);
        cost += a;
      } else {
        cost += b;
      }
    }
    d[i] = min(d[i], cost);
  }
  cout << d[n] << '\n';
}

```