from collections import deque

def check(mid): # mid より大きい辺を使う最小回数
    d = [1000000]*n
    d[0] = 0
    q = deque()
    q.append((0,0))
    while q:
        v,dist = q.popleft()
        if v==n-1:
            return d[v]
        if d[v] > dist: continue
        for u,c in g[v]:
            cost = int(c > mid)
            nd = d[v] + cost
            if nd < d[u]:
                d[u] = nd
                if cost: q.append((u,nd))
                else: q.appendleft((u,nd))
    return 1000000

import sys
readline = sys.stdin.readline
n,m,k = map(int,readline().split())
g = [[] for _ in range(n)]
for _ in range(m):
    u,v,c = map(int,readline().split())
    g[u-1].append((v-1,c))
    g[v-1].append((u-1,c))
ng = -1
ok = 2*10**5+1
while ok-ng>1:
    mid = (ok+ng)//2
    if check(mid) < k: ok = mid
    else: ng = mid
print(ok)