#include<bits/stdc++.h>
using namespace std;

#define int long long
const int N = 5e3 + 3;

int dp[N], A[N];

signed main(){
	int n, a, b;
	cin >> n >> a >> b;
    
    memset(dp , 22, sizeof dp);
    dp[0] = 0;
    for(int i=1; i<=n; i++){
        cin >> A[i];
        int t = 0;
        int mx = 0;
        for(int j=i-1; j>=0; j--){
            if(A[j] > A[i]) t++;
        	else mx = max(mx , A[j]);
            if(mx == A[j])
                dp[i] = min(dp[i] , dp[j] + a*t + b*(i-j-1-t));
        }
    }
    
    int mx = 0;
    int ans = 1LL << 60;
    for(int i=n; i>=1; i--){
        if(mx < A[i]){
            mx = A[i];
            ans = min(ans , dp[i] + (n-i)*b);
        }
    }
    
    cout << ans;
    return 0;
}
### Prompt

Your challenge is to write 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;

#define int long long
const int N = 5e3 + 3;

int dp[N], A[N];

signed main(){
	int n, a, b;
	cin >> n >> a >> b;
    
    memset(dp , 22, sizeof dp);
    dp[0] = 0;
    for(int i=1; i<=n; i++){
        cin >> A[i];
        int t = 0;
        int mx = 0;
        for(int j=i-1; j>=0; j--){
            if(A[j] > A[i]) t++;
        	else mx = max(mx , A[j]);
            if(mx == A[j])
                dp[i] = min(dp[i] , dp[j] + a*t + b*(i-j-1-t));
        }
    }
    
    int mx = 0;
    int ans = 1LL << 60;
    for(int i=n; i>=1; i--){
        if(mx < A[i]){
            mx = A[i];
            ans = min(ans , dp[i] + (n-i)*b);
        }
    }
    
    cout << ans;
    return 0;
}
```