#include<bits/stdc++.h>
using namespace std;
int pos[5000];
long long dp[6000][6000];
int main(){
  int n;
  long long a,b;
  scanf("%d%lld%lld",&n,&b,&a);
  for(int i=0;i<n;i++){
    int p;
    scanf("%d",&p);
    pos[p]=i+1;
  }
  for(int i=0;i<n+1;i++){
      if(i<pos[1])
        dp[1][i]=a;
      else
        dp[1][i]=0ll;
  }
  for(int x=2;x<=n;x++){
    for(int i=0;i<n+1;i++){
      dp[x][i] = (long long)1e18;
      if(i<pos[x])
        dp[x][i] = min(dp[x][i],a+dp[x-1][i]);
      else{
          dp[x][i] = min(dp[x][i], dp[x-1][pos[x]-1]);
          dp[x][i] = min(dp[x][i], b+dp[x-1][i]);
      }
    }
  }
  printf("%lld\n",dp[n][n]);
}

### 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<bits/stdc++.h>
using namespace std;
int pos[5000];
long long dp[6000][6000];
int main(){
  int n;
  long long a,b;
  scanf("%d%lld%lld",&n,&b,&a);
  for(int i=0;i<n;i++){
    int p;
    scanf("%d",&p);
    pos[p]=i+1;
  }
  for(int i=0;i<n+1;i++){
      if(i<pos[1])
        dp[1][i]=a;
      else
        dp[1][i]=0ll;
  }
  for(int x=2;x<=n;x++){
    for(int i=0;i<n+1;i++){
      dp[x][i] = (long long)1e18;
      if(i<pos[x])
        dp[x][i] = min(dp[x][i],a+dp[x-1][i]);
      else{
          dp[x][i] = min(dp[x][i], dp[x-1][pos[x]-1]);
          dp[x][i] = min(dp[x][i], b+dp[x-1][i]);
      }
    }
  }
  printf("%lld\n",dp[n][n]);
}

```