def oi(): return int(input())
def os(): return input()
def mi(): return list(map(int, input().split()))

# import sys
# input = sys.stdin.readline
# import sys
# sys.setrecursionlimit(10**8)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
input_count = 0

input_count = 0
N,M,K = mi()
NODE = N
LOOP_NUM = M
maps = {i:[] for i in range(NODE)}
cost_dict = {}
for i in range(LOOP_NUM):
    a,b,c = mi()
    maps[a-1].append((b-1, c, i))
    maps[b-1].append((a-1,c, i))
# # K 未満の時
# from collections import deque
# reached = [0] * N ## 1つ前のノードを持たせると微妙に遅くなるのでやめよう
# deq = deque([[0,0]])
# mins = float("inf")
# while deq:
#     node, count = deq.pop() #BFS

#     if node == N-1:
#         if count < K :
#             mins = 0
#             break

#     reached[node] = -1

#     for next_node, cost in maps[node]:
#         # 訪問済み判定
#         if reached[next_node] == -1:
#             continue
        
#         if count+1 <= K:
#             mins = min(mins, cost)
#         reached[next_node] = -1
#         deq.append([next_node, count+1])

# 以下の問題にてテスト済み
# https://atcoder.jp/contests/abc213/tasks/abc213_e
from collections import deque
def zero_one_dfs(target, K):
    zero_deq = deque([0])
    count=0
    use_node = set([])

    while True:
        one_deq = deque([])
        
        while zero_deq:
            node = zero_deq.pop()
                
            # 目的地に到達したら
            if node == N-1:
                if count < K:
                    return True

            # コスト0の時の移動パターン
            for next_node, cost, ind in maps[node]:
                if ind not in use_node:
                    use_node.add(ind)
                    if cost > target:
                        one_deq.append(next_node)
                    else:
                        zero_deq.append(next_node)

        count+=1
        zero_deq = one_deq
        if count >= K:
            return False
def is_ok(arg):
    # 条件を満たすかどうか？問題ごとに定義
    # 初めてFalseからTrueになるTrueの地点を取得
    return zero_one_dfs(arg, K)

def meguru_bisect(ng, ok):
    '''
    初期値のng,okを受け取り,is_okを満たす最小(最大)のokを返す
    まずis_okを定義すべし
    ng ok は  とり得る最小の値-1 とり得る最大の値+1
    最大最小が逆の場合はよしなにひっくり返す
    '''
    while (abs(ok - ng) > 1):
        mid = (ok + ng) // 2
        if is_ok(mid):
            ok = mid
        else:
            ng = mid
    return ok
# if mins != 0:
    
mins = meguru_bisect(-1, 2*10**5 + 1)
print(mins)