/*
二分答案
若 Li > mid, Ri = 1；否则 Ri = 0.
check():在以 R 为边权的新图中 1 至 N 的最短路若<=K，返回 1；否则返回 0

时间复杂度：
logL * NlogN=20*1000*10=200000
*/

#include <bits/stdc++.h>
using namespace std;
const int N = 1500, M = 10050;
typedef pair<int, int> pa;
struct Edge {
  int u, v, w;
} e[M];
int n, m, k, R[M], di[N], bk[N];
vector<pa> E[N];
inline int getDi() {
  for (int i = 1; i <= n; ++i) {
    E[i].clear();
  }
  for (int i = 1; i <= m; ++i) {
    E[e[i].u].push_back({e[i].v, R[i]});
    E[e[i].v].push_back({e[i].u, R[i]});
  }
  memset(bk, 0, sizeof bk);
  memset(di, 0x3f, sizeof di); di[1] = 0;
  priority_queue<pa, vector<pa>, greater<pa> > q;
  q.push({0, 1});
  while (q.size()) {
    int u = q.top().second; q.pop();
    if (bk[u]) continue;
    else bk[u] = 1;
    for (int i = 0; i < E[u].size(); ++i) {
      int v = E[u][i].first, w = E[u][i].second;
      if (di[v] > di[u] + w) q.push({di[v] = di[u] + w, v});
    }
  }
  return di[n];
}
inline bool check(int co) {
  for (int i = 1; i <= m; ++i)
    if (e[i].w > co) R[i] = 1;
    else R[i] = 0;
  int res = getDi();
  if (res <= k) return 1;
  return 0;
}
signed main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i)
    cin >> e[i].u >> e[i].v >> e[i].w;
  int l = 0, r = 1e6 + 5, mid;
  while (l < r) {
    mid = l + r >> 1;
    if (check(mid)) r = mid;
    else l = mid + 1;
  }
  if (l == 1e6 + 5) puts("-1");
  else cout << l;
  return 0;
}