#include <bits/stdc++.h>

using namespace std;

const int N = 5050;
const long long inf = 1e18;

long long dp[N][N];
int p[N];

void upd(long long &a, long long b) {
  if (a > b)
    a = b;
}

int main() {
  int n, a, b;
  scanf("%d %d %d", &n, &a, &b);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p+i);
  }
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j)
      dp[i][j] = inf;
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (dp[i][j] >= inf) continue;
      upd(dp[i+1][j], dp[i][j] + (j > p[i] ? b : a));
      if (j < p[i])
        upd(dp[i+1][p[i]], dp[i][j]);
    }
  }
  long long ans = inf;
  for (int j = 0; j <= n; ++j)
    upd(ans, dp[n][j]);
  printf("%lld\n", ans);
  return 0;
}
### 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;

const int N = 5050;
const long long inf = 1e18;

long long dp[N][N];
int p[N];

void upd(long long &a, long long b) {
  if (a > b)
    a = b;
}

int main() {
  int n, a, b;
  scanf("%d %d %d", &n, &a, &b);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p+i);
  }
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j)
      dp[i][j] = inf;
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (dp[i][j] >= inf) continue;
      upd(dp[i+1][j], dp[i][j] + (j > p[i] ? b : a));
      if (j < p[i])
        upd(dp[i+1][p[i]], dp[i][j]);
    }
  }
  long long ans = inf;
  for (int j = 0; j <= n; ++j)
    upd(ans, dp[n][j]);
  printf("%lld\n", ans);
  return 0;
}
```