#include <iostream>
using namespace std;
const int maxn=5e3+2;
long long n, A, B, dp[maxn], a[maxn];

void Enter()
{
    cin>>n>>A>>B;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    a[n+1]=1e18;
    for(int i=1;i<=n+1;i++)
    {
        dp[i]=1e18;
        long long t=0;
        for(int j=i-1;j>=0;j--)
        {
            if(a[j]>a[i])t++;
            else
                dp[i]=min(dp[i],dp[j]+t*A+(i-j-1-t)*B);
        }
    }
    cout<<dp[n+1];
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    Enter();
}

### 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 <iostream>
using namespace std;
const int maxn=5e3+2;
long long n, A, B, dp[maxn], a[maxn];

void Enter()
{
    cin>>n>>A>>B;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    a[n+1]=1e18;
    for(int i=1;i<=n+1;i++)
    {
        dp[i]=1e18;
        long long t=0;
        for(int j=i-1;j>=0;j--)
        {
            if(a[j]>a[i])t++;
            else
                dp[i]=min(dp[i],dp[j]+t*A+(i-j-1-t)*B);
        }
    }
    cout<<dp[n+1];
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    Enter();
}

```