# 我嫌每次输出都要写05d麻烦，直接用字典存了
e = {};ne = {}

h,n,k = input().split()
k = int(k)

for i in range(int(n)):
    addr,data,next = input().split()
    e[addr] = int(data)
    ne[addr] = next

"""
因为题干说明了每一类内部元素的顺序是不能改变的，将该链表划分为三个部分
遍历该链表的每一个元素，分别和0和k比较，将其划分到相应的part
"""
part1 = []
part2 = []
part3 = []

i = h
while i != "-1":
    val = e[i]
    if val < 0:
        part1.append([i,e[i]])
    elif val >= 0 and val <= k:
        part2.append([i,e[i]])
    else:
        part3.append([i,e[i]])
    i = ne[i]

parts = [part1,part2,part3]
ans = []

for part in parts: # 将三个部分合并
    for x in part:
        ans.append(x)

for i in range(len(ans)):
    if i == len(ans) - 1:
        print(ans[i][0],ans[i][1],-1)
    else:
        print(ans[i][0],ans[i][1],ans[i + 1][0])