#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
  int n;
  ll a, b;
  scanf("%d%lld%lld", &n, &a, &b);
  vector<ll> f(n + 1, 1ll << 55);
  f[0] = 0;
  for (int _ = 1; _ <= n; _++) {
    int x;
    scanf("%d", &x);
    vector<ll> g(n + 1, 1ll << 55);
    for (int i = 0; i <= n; i++) if (f[i] != 1ll << 55) {
      if (x > i) {
        g[x] = min(g[x], f[i]);
        g[i] = min(g[i], f[i] + a);
      } else {
        g[i] = min(g[i], f[i] + b);
      }
    }
    f = g;
  }
  ll ans = f[0];
  for (int i = 1; i <= n; i++) {
    ans = min(ans, f[i]);
  }
  printf("%lld\n", ans);
}


### Prompt

Please formulate 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;
typedef long long ll;
int main() {
  int n;
  ll a, b;
  scanf("%d%lld%lld", &n, &a, &b);
  vector<ll> f(n + 1, 1ll << 55);
  f[0] = 0;
  for (int _ = 1; _ <= n; _++) {
    int x;
    scanf("%d", &x);
    vector<ll> g(n + 1, 1ll << 55);
    for (int i = 0; i <= n; i++) if (f[i] != 1ll << 55) {
      if (x > i) {
        g[x] = min(g[x], f[i]);
        g[i] = min(g[i], f[i] + a);
      } else {
        g[i] = min(g[i], f[i] + b);
      }
    }
    f = g;
  }
  ll ans = f[0];
  for (int i = 1; i <= n; i++) {
    ans = min(ans, f[i]);
  }
  printf("%lld\n", ans);
}


```