思路：

1. 首先计算2路时的前n
2. 然后根据第i路的前n，加上第i+1路的数据，作为新的2路前n进行计算

2路前n的计算方法：

a 排序后，ax + by 的前 i 必然位于下面的候选项里：（b非排序，如果排序则最多到bi)

a0 + b0, a1 + b0, a2 + b0, …
a0 + b1, a1 + b1, a2 + b1, …
…
a0 + bi, a1 + bi, a2 + bi, …

from heapq import * 

N = 2010 

a, b, c = [[0] * N for _ in range(3)]

def merge():
    global a, b, c
    q = []
    for i in range(n):
        heappush(q, (b[i] + a[0], 0))

    for i in range(n):
        val, ai = heappop(q)
        c[i] = val 
        if ai < n - 1: 
            heappush(q, (val - a[ai] + a[ai + 1], ai + 1))

    a = list(c)

t = int(input())

while t:
    t -= 1

    m, n = map(int, input().split())

    a = sorted(map(int, input().split())) 

    for i in range(m - 1):
        b = list(map(int, input().split()))
        merge()

    for i in range(n):
        print(a[i], end=' ')
    print('')