#include <cstdio>
#include <algorithm>
#include <stdint.h>

int ps[5005];
int64_t dp[5005];

int main(){
  int N,A,B;
  scanf("%d %d %d",&N,&A,&B);
  for(int i=1;i<=N;i++){
    scanf("%d",&ps[i]);
    for(int j=0;j<=N;j++){
      if(j<ps[i]){
	dp[j]+=A;
      }else if(j>ps[i]){
	dp[j]+=B;
      }
    }
    for(int j=1;j<=N;j++){
      dp[j]=std::min(dp[j],dp[j-1]);
    }
  }
  printf("%ld\n",dp[N]);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <cstdio>
#include <algorithm>
#include <stdint.h>

int ps[5005];
int64_t dp[5005];

int main(){
  int N,A,B;
  scanf("%d %d %d",&N,&A,&B);
  for(int i=1;i<=N;i++){
    scanf("%d",&ps[i]);
    for(int j=0;j<=N;j++){
      if(j<ps[i]){
	dp[j]+=A;
      }else if(j>ps[i]){
	dp[j]+=B;
      }
    }
    for(int j=1;j<=N;j++){
      dp[j]=std::min(dp[j],dp[j-1]);
    }
  }
  printf("%ld\n",dp[N]);
  return 0;
}

```