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#include <bits/stdc++.h>
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using namespace std;
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typedef pair<int, int> PII;
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const int INF = 0x3f3f3f3f;
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const int N = 1010; // 1000个点
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const int M = 20010; // 10000条，记录无向边需要两倍空间
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int idx, h[N], e[M], w[M], ne[M];
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void add(int a, int b, int c) {
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e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
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}
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int n; //点数
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int m; //边数
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bool st[N]; //记录是不是在队列中
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int k; //不超过K条电缆，由电话公司免费提供升级服务
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int dist[N]; //记录最短距离
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// u指的是我们现在选最小花费
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bool check(int x) {
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memset(st, false, sizeof st);
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memset(dist, 0x3f, sizeof dist);
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priority_queue<PII, vector<PII>, greater<PII>> q;
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dist[1] = 0;
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q.push({0, 1});
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while (q.size()) {
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PII t = q.top();
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q.pop();
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int d = t.first, u = t.second;
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if (st[u]) continue;
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st[u] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i], v = w[i] > x; //如果有边比我们现在选的这条边大，那么这条边对方案的贡献为1，反之为0
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if (dist[j] > d + v) {
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dist[j] = d + v;
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q.push({dist[j], j});
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}
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}
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}
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//如果按上面的方法计算后，n结点没有被松弛操作修改距离，则表示n不可达
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if (dist[n] == INF) {
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puts("-1"); //不可达，直接输出-1
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exit(0);
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}
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return dist[n] <= k; //如果有k+1条边比我们现在这条边大，那么这个升级方案就是不合法的，反之就合法
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}
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int main() {
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memset(h, -1, sizeof h);
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cin >> n >> m >> k;
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int a, b, c;
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for (int i = 0; i < m; i++) {
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cin >> a >> b >> c;
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add(a, b, c), add(b, a, c);
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}
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/*这里二分的是直接面对答案设问：最少花费
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依题意，最少花费其实是所有可能的路径中，第k+1条边的花费
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如果某条路径不存在k+1条边(边数小于k+1),此时花费为0
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同时，任意一条边的花费不会大于1e6
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整理一下，这里二分枚举的值其实是0 ~ 1e6*/
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int l = 0, r = 1e6;
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while (l < r) {
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int mid = (l + r) >> 1;
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if (check(mid)) // check函数的意义：如果当前花费可以满足要求，那么尝试更小的花费
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r = mid;
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else
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l = mid + 1;
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}
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printf("%d\n", l);
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return 0;
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}
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