XxOoRr.cpp

/* 
    Solution by Rahul Surana
    
    ***********************************************************


Given an array A1,A2…AN, find the minimum number of operations (possibly zero) required to convert all integers in A to 0.

In one operation, you

choose a non-negative integer p (p≥0),
select at most K indices in the array A, and
for each selected index i, replace Ai with Ai⊕2p. Here, ⊕ denotes bitwise XOR.

Input

The first line contains an integer T - the number of test cases. Then T test cases follow.
The first line of each test case contains two integers N, K - the size of the array and the maximum number of elements you can select in an operation.
The second line of each test case contains N integers A1,A2…AN.

Output

For each test case, output the minimum number of operations to make all elements of the array 0.



    ***********************************************************
*/


#include <bits/stdc++.h>
#define ll          long long
#define vl          vector<ll>
#define vi          vector<int>
#define pi          pair<int,int>
#define pl          pair<ll,ll>
#define all(a)      a.begin(),a.end()
#define mem(a,x)    memset(a,x,sizeof(a))
#define pb          push_back
#define mp          make_pair
#define F           first
#define S           second
#define FOR(i,a)     for(int i = 0; i < a; i++)
#define trace(x)    cerr<<#x<<" : "<<x<<endl;
#define trace2(x,y) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<endl;
#define trace3(x,y,z) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<" | "<<#z<<" : "<<z<<endl;
#define fast_io 	std::ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
 
using namespace std;

int ar[33];

int main()
{
	fast_io;
    int t;
    cin >> t;
    while(t--) {
        int n,k;
        cin >> n >> k;
        FOR(i,33) ar[i]=0;
        FOR(i,n){
            int x;
            cin >> x;
            int j = 0;
            while(x>0){
                if(x%2) ar[j]++;
                j++;
                x = x>>1;
            }
        }
        ll ans = 0;
        for(int i = 0; i < 33;i++){
            ans+= ar[i]/k + (ar[i]%k > 0);
        }
        cout << ans<<"\n";
    }
}