/*
 * 2536.cc: Median Tree
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 1000;
const int MAX_M = 10000;

/* typedef */

struct UFT {
  int links[MAX_N], ranks[MAX_N], sizes[MAX_N];

  void clear(int n) {
    for (int i = 0; i < n; i++) links[i] = i, ranks[i] = sizes[i] = 1;
  }

  UFT() {}
  UFT(int n) { clear(n); }

  int root(int i) {
    int i0 = i;
    while (links[i0] != i0) i0 = links[i0];
    return (links[i] = i0);
  }

  int rank(int i) { return ranks[root(i)]; }
  int size(int i) { return sizes[root(i)]; }
  bool same(int i, int j) { return root(i) == root(j); }

  int merge(int i0, int i1) {
    int r0 = root(i0), r1 = root(i1), mr;
    if (ranks[r0] == ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      ranks[r0]++;
      mr = r0;
    }
    else if (ranks[r0] > ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      mr = r0;
    }
    else {
      links[r0] = r1;
      sizes[r1] += sizes[r0];
      mr = r1;
    }
    return mr;
  }
};

struct Edge {
  int u, v, c;
  Edge() {}
  Edge(int _u, int _v, int _c): u(_u), v(_v), c(_c) {}
  bool operator<(const Edge &e) const { return c < e.c; }
};

/* global variables */

UFT uft;
Edge es[MAX_M];

/* subroutines */

/* main */

int main() {
  for (;;) {
    int n, m;
    cin >> n >> m;
    if (n == 0) break;

    for (int i = 0; i < m; i++) {
      cin >> es[i].u >> es[i].v >> es[i].c;
      es[i].u--, es[i].v--;
    }
    sort(es, es + m);

    uft.clear(n);
    int en = 0, me = (n - 1) / 2, ans = -1;

    for (int i = 0; i < m; i++) {
      Edge &ei = es[i];
      int ru = uft.root(ei.u);
      int rv = uft.root(ei.v);
      if (ru != rv) {
	uft.merge(ei.u, ei.v);
	if (en == me) {
	  ans = ei.c;
	  break;
	}
	en++;
      }
    }

    printf("%d\n", ans);
  }
  return 0;
}
### Prompt

In CPP, your task is to solve the following problem:

Problem Statement

You are given a connected undirected graph which has even numbers of nodes. A connected graph is a graph in which all nodes are connected directly or indirectly by edges.

Your task is to find a spanning tree whose median value of edges' costs is minimum. A spanning tree of a graph means that a tree which contains all nodes of the graph.



Input

The input consists of multiple datasets.

The format of each dataset is as follows.


n m
s_1 t_1 c_1
...
s_m t_m c_m


The first line contains an even number n (2 \leq n \leq 1,000) and an integer m (n-1 \leq m \leq 10,000). n is the nubmer of nodes and m is the number of edges in the graph.

Then m lines follow, each of which contains s_i (1 \leq s_i \leq n), t_i (1 \leq s_i \leq n, t_i \neq s_i) and c_i (1 \leq c_i \leq 1,000). This means there is an edge between the nodes s_i and t_i and its cost is c_i. There is no more than one edge which connects s_i and t_i.

The input terminates when n=0 and m=0. Your program must not output anything for this case.

Output

Print the median value in a line for each dataset.

Example

Input

2 1
1 2 5
4 6
1 2 1
1 3 2
1 4 3
2 3 4
2 4 5
3 4 6
8 17
1 4 767
3 1 609
8 3 426
6 5 972
8 1 607
6 4 51
5 1 683
3 6 451
3 4 630
8 7 912
3 7 43
4 7 421
3 5 582
8 4 538
5 7 832
1 6 345
8 2 608
0 0


Output

5
2
421

### Response

```cpp
/*
 * 2536.cc: Median Tree
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 1000;
const int MAX_M = 10000;

/* typedef */

struct UFT {
  int links[MAX_N], ranks[MAX_N], sizes[MAX_N];

  void clear(int n) {
    for (int i = 0; i < n; i++) links[i] = i, ranks[i] = sizes[i] = 1;
  }

  UFT() {}
  UFT(int n) { clear(n); }

  int root(int i) {
    int i0 = i;
    while (links[i0] != i0) i0 = links[i0];
    return (links[i] = i0);
  }

  int rank(int i) { return ranks[root(i)]; }
  int size(int i) { return sizes[root(i)]; }
  bool same(int i, int j) { return root(i) == root(j); }

  int merge(int i0, int i1) {
    int r0 = root(i0), r1 = root(i1), mr;
    if (ranks[r0] == ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      ranks[r0]++;
      mr = r0;
    }
    else if (ranks[r0] > ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      mr = r0;
    }
    else {
      links[r0] = r1;
      sizes[r1] += sizes[r0];
      mr = r1;
    }
    return mr;
  }
};

struct Edge {
  int u, v, c;
  Edge() {}
  Edge(int _u, int _v, int _c): u(_u), v(_v), c(_c) {}
  bool operator<(const Edge &e) const { return c < e.c; }
};

/* global variables */

UFT uft;
Edge es[MAX_M];

/* subroutines */

/* main */

int main() {
  for (;;) {
    int n, m;
    cin >> n >> m;
    if (n == 0) break;

    for (int i = 0; i < m; i++) {
      cin >> es[i].u >> es[i].v >> es[i].c;
      es[i].u--, es[i].v--;
    }
    sort(es, es + m);

    uft.clear(n);
    int en = 0, me = (n - 1) / 2, ans = -1;

    for (int i = 0; i < m; i++) {
      Edge &ei = es[i];
      int ru = uft.root(ei.u);
      int rv = uft.root(ei.v);
      if (ru != rv) {
	uft.merge(ei.u, ei.v);
	if (en == me) {
	  ans = ei.c;
	  break;
	}
	en++;
      }
    }

    printf("%d\n", ans);
  }
  return 0;
}
```