Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=5005;
long long s[N],p[N],f[N];
int n,A,B;
int main(){
    freopen("1.out","w",stdout);
    freopen("1.in","r",stdin);
    memset(f,23,sizeof(f));
    scanf("%d%d%d",&n,&A,&B),f[0]=0;
    for(int i=1,x;i<=n;i++){
        scanf("%d",&x);
        for(int j=0;j<=n;j++)
            if(j)   p[j]=min(p[j-1],f[j]);
            else    p[j]=f[j];
        for(int j=0;j<=x;j++)   f[j]+=A;
        for(int j=x+1;j<=n;j++) f[j]+=B;
        f[x]=min(f[x],p[x]);
    }
    long long ans=f[n+1];
    for(int i=0;i<=n;i++)   ans=min(ans,f[i]);
    printf("%lld\n",ans);
    return 0;
}