codeforces 985A Chess Placing

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本文介绍了一个关于在1×n的黑白相间棋盘上，将n/2个棋子移动到同一颜色格子上的最少移动次数的问题。通过输入棋子的初始位置，算法计算出将所有棋子移动到相同颜色格子所需的最小步骤。

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You are given a chessboard of size 1 × n. It is guaranteed that n is even. The chessboard is painted like this: "BWBW...BW".

Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to $\text{[math]}$ .

In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.

Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100, n is even) — the size of the chessboard.

The second line of the input contains $\text{[math]}$ integer numbers $\text{[math]}$ (1 ≤ p_i ≤ n) — initial positions of the pieces. It is guaranteed that all the positions are distinct.

Output

Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color.

Examples

Input

Copy

6
1 2 6

Output

Copy

Input

Copy

10
1 2 3 4 5

Output

Copy

Note

In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2 to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.

In the second example the possible strategy is to move $\text{[math]}$ in 4 moves, then $\text{[math]}$ in 3 moves, $\text{[math]}$ in 2 moves and $\text{[math]}$ in 1 move.

题意

n大小的黑白棋盘，上面有n/2个棋子，告诉你位置，问最少移动多少次使他们全部在白色或全部在黑色

思路

分别移动到白色和黑色，看那个步数最少

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int a[120],b[120];
int main()
{
    int n,i,j,x,sum1,sum2;
    while(cin>>n)
    {
        for(i=0;i<n/2;i++)
        {
            cin>>x;
            a[x] = 1;
            b[x] = 1;
        }
        x = 1;
        sum1 = 0;sum2 = 0;
        for(i=1;i<=n;i++)
        {
            if(a[i]==1)
            {
                sum1 += abs(x-i);
                x += 2;
            }
        }
        x=2;
        for(i=1;i<=n;i++)
        {
            if(a[i]==1)
            {
                sum2 += abs(x-i);
                x += 2;
            }
        }
        if(sum1>sum2) sum1 = sum2;
        cout<<sum1<<endl;
    }
    return 0;
}