#include <cstdio>
#include <vector>
using namespace std;

const long long INF=10000000000000000ll;

int main()
{
  int n;
  long long a,b;
  scanf("%d%lld%lld",&n,&a,&b);
  vector<int> p(n+1);
  for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
    p[i]--;
  }
  p[n]=n;
  vector<long long> dp(n+1);
  for(int i=0;i<=n;i++){
    int ca=0,cb=0;
    int k=-1;
    dp[i]=INF;
    for(int j=i-1;j>=0;j--){
      if(p[j]>p[i]){
	ca++;
      }
      else if(k==-1||p[k]<p[j]){
	dp[i]=min(dp[i],dp[j]+ca*a+cb*b);
	cb++;
	k=j;
      }
      else{
	cb++;
      }
    }
    if(k==-1){
      dp[i]=i*a;
    }
  }
  printf("%lld\n",dp[n]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <vector>
using namespace std;

const long long INF=10000000000000000ll;

int main()
{
  int n;
  long long a,b;
  scanf("%d%lld%lld",&n,&a,&b);
  vector<int> p(n+1);
  for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
    p[i]--;
  }
  p[n]=n;
  vector<long long> dp(n+1);
  for(int i=0;i<=n;i++){
    int ca=0,cb=0;
    int k=-1;
    dp[i]=INF;
    for(int j=i-1;j>=0;j--){
      if(p[j]>p[i]){
	ca++;
      }
      else if(k==-1||p[k]<p[j]){
	dp[i]=min(dp[i],dp[j]+ca*a+cb*b);
	cb++;
	k=j;
      }
      else{
	cb++;
      }
    }
    if(k==-1){
      dp[i]=i*a;
    }
  }
  printf("%lld\n",dp[n]);
  return 0;
}

```