#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n; cin >> n;
  lint A, B; cin >> A >> B;
  V<> p(n); for (auto&& e : p) cin >> e, --e;
  V<lint> dp(n + 1);
  for (int x = 0; x < n; ++x) {
    int j = find(begin(p), end(p), x) - begin(p);
    int c = 0;
    for (int i = j + 1; ; ++i) {
      if (c) dp[i] += A;
      dp[i] = min(dp[i], B * c + dp[j]);
      if (i == n) break;
      c += p[i] < x;
    }
  }
  cout << dp.back() << '\n';
}
### 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>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n; cin >> n;
  lint A, B; cin >> A >> B;
  V<> p(n); for (auto&& e : p) cin >> e, --e;
  V<lint> dp(n + 1);
  for (int x = 0; x < n; ++x) {
    int j = find(begin(p), end(p), x) - begin(p);
    int c = 0;
    for (int i = j + 1; ; ++i) {
      if (c) dp[i] += A;
      dp[i] = min(dp[i], B * c + dp[j]);
      if (i == n) break;
      c += p[i] < x;
    }
  }
  cout << dp.back() << '\n';
}
```