Maximize sum after K negations

PROBLEM :
Given an array of size n and a number k. We must modify array K number of times. Here modify array means in each operation we can replace any array element arr[i] by -arr[i]. We need to perform this operation in such a way that after K operations, sum of array must be maximum?

Examples:

Input : arr[] = {-2, 0, 5, -1, 2}
        K = 4
Output: 10
// Replace (-2) by -(-2), array becomes {2, 0, 5, -1, 2}
// Replace (-1) by -(-1), array becomes {2, 0, 5, 1, 2}
// Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}
// Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}

Input : arr[] = {9, 8, 8, 5}
        K = 3
Output: 20
Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consist of three lines . The first line of each test case contains an integer N.The second line of each test case contains N space separated integers denoting elements of the array. Third line contains value of k.

Output:
For each test case in a new line print maximum possible sum.

Constraints:
1 = T = 500
1 = N = 1000

Example:
Input:
2
5
1 2 -3 4 5
1
10
5 -2 5 -4 5 -12 5 5 5 20
5

Output:
15
68

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
#define max 1000
int Maximize_sum(int [],int ,int ) ;
int main()
 {
    int t,no,i,k ;
    int arr[max] ;
    cin>>t ;
    while(t--)
    {
        cin>>no ;
        for(i=0;i<no;i++)
            cin>>arr[i] ;
        cin>>k ;
        cout<<Maximize_sum(arr,no,k)<<endl ;
    }
return 0;
}

int Maximize_sum(int arr[],int no,int k)
{
    int i,min ;
   
    while(k--)
    {
        min=0 ;
        for(i=0;i<no;i++)
            if(arr[i]<arr[min])
                min=i ;
               
        if(arr[min]==0)
            break ;
           
        arr[min]=-arr[min] ;
    }
   
    int sum=0 ;
   
    for(i=0;i<no;i++)
        sum=sum+arr[i] ;
       
    return sum ;
}

---------------------------------------------------------------------------------

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