#include <bits/stdc++.h>
using namespace std;
int n;
long long AA, BB;
int t[5009];
long long dp[5009][5009]; //from i, limit j
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>AA>>BB;
    for(int i=1; i<=n; i++)
    {
        cin>>t[i];
    }
    for(int i=n; i>=1; i--)
    {
        for(int lim=0; lim<=n; lim++)
        {
            if(t[i]<lim)
            {
                dp[i][lim]=dp[i+1][lim]+BB;
            }
            else
            {
                dp[i][lim]=min(dp[i+1][t[i]], dp[i+1][lim]+AA);
            }
        }
    }
    cout<<dp[1][0]<<endl;
    return 0;
}

### Prompt

Please formulate 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>
using namespace std;
int n;
long long AA, BB;
int t[5009];
long long dp[5009][5009]; //from i, limit j
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>AA>>BB;
    for(int i=1; i<=n; i++)
    {
        cin>>t[i];
    }
    for(int i=n; i>=1; i--)
    {
        for(int lim=0; lim<=n; lim++)
        {
            if(t[i]<lim)
            {
                dp[i][lim]=dp[i+1][lim]+BB;
            }
            else
            {
                dp[i][lim]=min(dp[i+1][t[i]], dp[i+1][lim]+AA);
            }
        }
    }
    cout<<dp[1][0]<<endl;
    return 0;
}

```