#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=5050;
const ll INF=1e15;
ll a[maxn][maxn],dp[maxn],p[maxn],n,A,B;

int main(){
    cin >> n >> A >> B;
    for (int i=1;i<=n;i++) cin >> p[i];
    p[++n]=n+1;
    for (int i=0;i<=n;i++){
        a[i][i]=0;
        for (int j=i+1;j<=n;j++) a[i][j]=a[i][j-1]+(p[j]>p[i]?A:B);
    }
    for (int i=1;i<=n;i++){
        dp[i]=INF;
        for (int j=0;j<i;j++) if (p[j]<p[i]) dp[i]=min(dp[i],dp[j]+a[j][i-1]);
    }
    cout << dp[n] << endl;
    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>
using namespace std;
typedef long long ll;
const ll maxn=5050;
const ll INF=1e15;
ll a[maxn][maxn],dp[maxn],p[maxn],n,A,B;

int main(){
    cin >> n >> A >> B;
    for (int i=1;i<=n;i++) cin >> p[i];
    p[++n]=n+1;
    for (int i=0;i<=n;i++){
        a[i][i]=0;
        for (int j=i+1;j<=n;j++) a[i][j]=a[i][j-1]+(p[j]>p[i]?A:B);
    }
    for (int i=1;i<=n;i++){
        dp[i]=INF;
        for (int j=0;j<i;j++) if (p[j]<p[i]) dp[i]=min(dp[i],dp[j]+a[j][i-1]);
    }
    cout << dp[n] << endl;
    return 0;
}

```