只考虑两组情况就行 然后依次迭代 a是有序的 
a1, a2, a3,……, an;

b1, b2, b3,……, bn;

import java.io.*;
import java.util.*;

class Main{
    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static int nextInt()throws Exception{
        in.nextToken();
        return (int)in.nval;
    }
    static class Pair{
        int first, second;
        public Pair(int first, int second){
            this.first = first;
            this.second = second;
        }
    }

    static int[] a, b, c;
    static int n, m;

    static void merge(){
        PriorityQueue<Pair> heap = new PriorityQueue<>((o1, o2)->o1.second-o2.second);
        for(int i=0; i<n; i++) heap.offer(new Pair(0, a[0]+b[i])); 
        for(int i=0; i<n; i++){
            Pair e = heap.poll();
            int f = e.first; int s = e.second;
            c[i] = s; 
            if(f<n-1)heap.offer(new Pair(f+1, s - a[f] + a[f+1])); 
        }
        for(int i=0; i<n; i++) a[i] = c[i];
    }

    public static void main(String[] args)throws Exception{
        int T = nextInt();
        while(T--!=0){
            m = nextInt(); n = nextInt();
            a = new int[n]; b = new int[n]; c = new int[n];
            for(int i=0; i<n; i++)a[i] = nextInt();
            Arrays.sort(a);
            for(int i=0; i<m-1; i++){
                for(int j=0; j<n; j++)b[j] = nextInt();
                merge(); 
            } 
            for(int i=0; i<n; i++) out.print(a[i]+" ");
            out.println("");
        }
        out.close();
    }
}