#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
#define pa pair<int,int>
const int N=1e5+10;
const int mod=1e9+7;
using namespace std;

int a[5005];
ll dp[5005];
int main()
{   
    ll n,A,B; scanf("%lld%lld%lld",&n,&A,&B);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    a[n+1]=n+1;
  for(int i=1;i<=n+1;++i)
  {
     dp[i]=1e18;
     int big=0;
     for(int j=i-1;j>=0;j--) 
     {
       if(a[i]<a[j]) big++;
       else dp[i]=min(dp[i],dp[j]+A*big+B*(i-1-j-big));
     }
  }
  printf("%lld\n",dp[n+1]);
return 0;
}
/*
*/
### Prompt

Your task is to create 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>
#define ll long long
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
#define pa pair<int,int>
const int N=1e5+10;
const int mod=1e9+7;
using namespace std;

int a[5005];
ll dp[5005];
int main()
{   
    ll n,A,B; scanf("%lld%lld%lld",&n,&A,&B);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    a[n+1]=n+1;
  for(int i=1;i<=n+1;++i)
  {
     dp[i]=1e18;
     int big=0;
     for(int j=i-1;j>=0;j--) 
     {
       if(a[i]<a[j]) big++;
       else dp[i]=min(dp[i],dp[j]+A*big+B*(i-1-j-big));
     }
  }
  printf("%lld\n",dp[n+1]);
return 0;
}
/*
*/
```