# input
import sys
input = sys.stdin.readline
II = lambda : int(input())
MI = lambda : map(int, input().split())
LI = lambda : [int(a) for a in input().split()]
SI = lambda : input().rstrip()
LLI = lambda n : [[int(a) for a in input().split()] for _ in range(n)]
LSI = lambda n : [input().rstrip() for _ in range(n)]
MI_1 = lambda : map(lambda x:int(x)-1, input().split())
LI_1 = lambda : [int(a)-1 for a in input().split()]

mod = 998244353
inf = 1001001001001001001
ordalp = lambda s : ord(s)-65 if s.isupper() else ord(s)-97
ordallalp = lambda s : ord(s)-39 if s.isupper() else ord(s)-97
yes = lambda : print("Yes")
no = lambda : print("No")
yn = lambda flag : print("Yes" if flag else "No")

prinf = lambda ans : print(ans if ans < 1000001001001001001 else -1)
alplow = "abcdefghijklmnopqrstuvwxyz"
alpup = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
alpall = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
URDL = {'U':(-1,0), 'R':(0,1), 'D':(1,0), 'L':(0,-1)}
DIR_4 = [[-1,0],[0,1],[1,0],[0,-1]]
DIR_8 = [[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1],[-1,-1]]
DIR_BISHOP = [[-1,1],[1,1],[1,-1],[-1,-1]]
prime60 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]
sys.set_int_max_str_digits(0)
# sys.setrecursionlimit(10**6)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')

from collections import defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right
DD = defaultdict
BSL = bisect_left
BSR = bisect_right

"""
all not 0 のとき、すべて 0 で ok

そうではないとき「満たしてしまう条件」として追加する

l <= の条件がつくと考える

適当にまわしまくれば O(NM) では解けそう

これの高速化をしたい

毎回確認するのがわるい
"""

def solve():
    n, m = MI()
    
    que = [[] for i in range(n)]
    op = []
    nd = [0] * m # いまの バージョン
    
    
    a = [0] * n
    
    vv = 1
    
    st = []
    ok = [0] * m
    for i in range(m):
        x, y, z, v, l, r = MI()
        x -= 1
        y -= 1
        z -= 1
        op.append((x, y, z, v, l, r))  
        if v == 0:
            ok[i] = 1
            st.append(i)
        else:
            d = (v + 1) // 2
            nd[i] = vv
            heappush(que[x], (d, i, vv))
            heappush(que[y], (d, i, vv))
            vv += 1  
    # すでに確認したかどうか
    
    while st:
        i = st.pop()
        _, _, zz, _, l, _ = op[i]
        if l <= a[zz]: continue
        
        # z が更新されたとき
        a[zz] = l
        while que[zz] and que[zz][0][0] <= a[zz]:
            _, j, nvv = heappop(que[zz])
            if nd[j] != nvv: continue
            # if ok[j]: continue
            
            x, y, z, v, l, r = op[j]
            # 発動
            if a[x] + a[y] >= v:
                st.append(j)
                ok[j] = 1
            else:
                # 発動せず
                # これをすれば log
                d = (v - (a[x] + a[y]) + 1) // 2
                nd[j] = vv
                heappush(que[x], (a[x] + d, j, vv))
                heappush(que[y], (a[y] + d, j, vv))
                vv += 1
    
    
    # print(a)
    f = 1
    for x, y, z, v, l, r in op:
        if a[x] + a[y] >= v:
            if not (l <= a[z] <= r):
                f = 0
                break
    
    if f == 0:
        print(-1)
    else:
        print(*a)

t = II()
for i in range(t):
    solve()