#include <iostream>
#define nmax 5005
using std::cin;
using std::cout;
long long min(long long x, long long y)
{
    return (y < x) ? y : x;
}
int n, x[nmax+1], p[nmax+1];
long long dp[nmax+1][nmax*2+1], a, b;
int main()
{
    std::ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> a >> b;
    for(int i = 1; i <= n; ++i)
      cin >> p[i];
    for(int i = 1; i <= n; ++i)
      x[p[i]] = (i << 1) - 1;
    for(int i = 1; i <= n; ++i){
      dp[i][0] = dp[i - 1][0] + b;
      for(int j = 1; j <= n << 1; ++j)
        dp[i][j] = min(dp[i][j - 1], ((j < x[i]) ? b : ((j > x[i]) ? a : 0)) + ((j & 1) ? dp[i - 1][j - 1] : dp[i - 1][j]));
    }
    cout << dp[n][n * 2];
    return 0;
}

### Prompt

Please create a solution in CPP 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 <iostream>
#define nmax 5005
using std::cin;
using std::cout;
long long min(long long x, long long y)
{
    return (y < x) ? y : x;
}
int n, x[nmax+1], p[nmax+1];
long long dp[nmax+1][nmax*2+1], a, b;
int main()
{
    std::ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> a >> b;
    for(int i = 1; i <= n; ++i)
      cin >> p[i];
    for(int i = 1; i <= n; ++i)
      x[p[i]] = (i << 1) - 1;
    for(int i = 1; i <= n; ++i){
      dp[i][0] = dp[i - 1][0] + b;
      for(int j = 1; j <= n << 1; ++j)
        dp[i][j] = min(dp[i][j - 1], ((j < x[i]) ? b : ((j > x[i]) ? a : 0)) + ((j & 1) ? dp[i - 1][j - 1] : dp[i - 1][j]));
    }
    cout << dp[n][n * 2];
    return 0;
}

```