#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;

int main(){
    int N;
    ll A,B;
    cin >> N >> A >> B;
    ll inf = 1e18;
    vector<int> l(N+1,0),p(N+1,0);
    for(int i=1;i<=N;i++) cin >> p[i];
    vector<vector<ll>> dp(N+1,vector<ll>(N+1,inf));
    for(int i=1;i<=N;i++) for(int j=1;j<=N;j++){
        if(p[j]==i) break;
        if(p[j]>i) l[i]++;
    }
    for(int j=0;j<=N;j++) dp[0][j] = 0;
    for(int i=1;i<=N;i++) for(int j=0;j<=N;j++){
        if(j<=l[i]){
            dp[i][j] = min(dp[i][j],dp[i-1][j]+B);
            dp[i][l[i]] = min(dp[i][l[i]],dp[i-1][j]);
        }else dp[i][j-1] = min(dp[i][j-1],dp[i-1][j]+A);
    }
    cout << dp[N][0] << endl;
}
### 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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;

int main(){
    int N;
    ll A,B;
    cin >> N >> A >> B;
    ll inf = 1e18;
    vector<int> l(N+1,0),p(N+1,0);
    for(int i=1;i<=N;i++) cin >> p[i];
    vector<vector<ll>> dp(N+1,vector<ll>(N+1,inf));
    for(int i=1;i<=N;i++) for(int j=1;j<=N;j++){
        if(p[j]==i) break;
        if(p[j]>i) l[i]++;
    }
    for(int j=0;j<=N;j++) dp[0][j] = 0;
    for(int i=1;i<=N;i++) for(int j=0;j<=N;j++){
        if(j<=l[i]){
            dp[i][j] = min(dp[i][j],dp[i-1][j]+B);
            dp[i][l[i]] = min(dp[i][l[i]],dp[i-1][j]);
        }else dp[i][j-1] = min(dp[i][j-1],dp[i-1][j]+A);
    }
    cout << dp[N][0] << endl;
}
```