You are given a connected undirected weighted graph consisting of n vertices and m edges.

You need to print the k-th smallest shortest path in this graph (paths from the vertex to itself are not counted, paths from i to j and from j to i are counted as one).

More formally, if d is the matrix of shortest paths, where d_{i, j} is the length of the shortest path between vertices i and j (1 ≤ i < j ≤ n), then you need to print the k-th element in the sorted array consisting of all d_{i, j}, where 1 ≤ i < j ≤ n.

Input

The first line of the input contains three integers n, m and k (2 ≤ n ≤ 2 ⋅ 10^5, n - 1 ≤ m ≤ min\Big((n(n-1))/(2), 2 ⋅ 10^5\Big), 1 ≤ k ≤ min\Big((n(n-1))/(2), 400\Big) — the number of vertices in the graph, the number of edges in the graph and the value of k, correspondingly.

Then m lines follow, each containing three integers x, y and w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10^9, x ≠ y) denoting an edge between vertices x and y of weight w.

It is guaranteed that the given graph is connected (there is a path between any pair of vertices), there are no self-loops (edges connecting the vertex with itself) and multiple edges (for each pair of vertices x and y, there is at most one edge between this pair of vertices in the graph).

Output

Print one integer — the length of the k-th smallest shortest path in the given graph (paths from the vertex to itself are not counted, paths from i to j and from j to i are counted as one).

Examples

Input


6 10 5
2 5 1
5 3 9
6 2 2
1 3 1
5 1 8
6 5 10
1 6 5
6 4 6
3 6 2
3 4 5


Output


3


Input


7 15 18
2 6 3
5 7 4
6 5 4
3 6 9
6 7 7
1 6 4
7 1 6
7 2 1
4 3 2
3 2 8
5 3 6
2 5 5
3 7 9
4 1 8
2 1 1


Output


9
from __future__ import division, print_function
 
DEBUG = 0
 
import os, sys
from atexit import register
from io import BytesIO
import itertools
 
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange
 
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
 
if DEBUG:
    debug_print = print
else:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
    sys.stdout = BytesIO()
    register(lambda: os.write(1, sys.stdout.getvalue()))
 
    input = lambda: sys.stdin.readline().rstrip('\r\n')
    debug_print = lambda *x, **y: None
 
 
def input_as_list():
    return list(map(int, input().split()))
 
 
def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
 
 
def main():
    from heapq import heapify, heappop, heappush
 
    n, m, k = input_as_list()
 
    g = array_of(list, n)
    elist = []
    pq = []
 
    for _ in range(m):
        u, v, d = input_as_list()
        elist.append((d, u-1, v-1))
 
    elist.sort()
    elist = elist[:k]
 
    V = set()
    for d, u, v in elist:
        V.add(u)
        V.add(v)
        g[u].append((v, d))
        g[v].append((u, d))
 
    all_paths = []
 
    for v in V:
        pq = [(0, v)]
        ub = dict()
        done = set()
 
        while pq:
            d, x = heappop(pq)
 
            if x in done:
                continue
 
            debug_print(v, x, d)
            done.add(x)
            if d > 0 : all_paths.append(d)
 
            for u, du in g[x]:
                du += d
 
                if u in ub and du >= ub[u]:
                    continue
 
                heappush(pq, (du, u))
                ub[u] = du
 
    all_paths.sort()
    print(all_paths[2*k-1])
 
main()