#include <bits/stdc++.h>
using namespace std;
struct unionfind{
  vector<int> p;
  unionfind(int n){
    p = vector<int>(n, -1);
  }
  int root(int x){
    if (p[x] == -1){
      return x;
    } else {
      p[x] = root(p[x]);
      return p[x];
    }
  }
  bool same(int x, int y){
    return root(x) == root(y);
  }
  void unite(int x, int y){
    x = root(x);
    y = root(y);
    p[x] = y;
  }
};
int main(){
  while (1){
    int n, m;
    cin >> n >> m;
    if (n == 0 && m == 0){
      break;
    }
    vector<tuple<int, int, int>> E(m);
    for (int i = 0; i < m; i++){
      int s, t, c;
      cin >> s >> t >> c;
      s--;
      t--;
      E[i] = make_tuple(c, s, t);
    }
    sort(E.begin(), E.end());
    unionfind UF(n);
    int cnt = 0;
    for (int i = 0; i < m; i++){
      int c = get<0>(E[i]);
      int s = get<1>(E[i]);
      int t = get<2>(E[i]);
      if (!UF.same(s, t)){
        UF.unite(s, t);
        cnt++;
        if (cnt == n / 2){
          cout << c << endl;
          break;
        }
      }
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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
#include <bits/stdc++.h>
using namespace std;
struct unionfind{
  vector<int> p;
  unionfind(int n){
    p = vector<int>(n, -1);
  }
  int root(int x){
    if (p[x] == -1){
      return x;
    } else {
      p[x] = root(p[x]);
      return p[x];
    }
  }
  bool same(int x, int y){
    return root(x) == root(y);
  }
  void unite(int x, int y){
    x = root(x);
    y = root(y);
    p[x] = y;
  }
};
int main(){
  while (1){
    int n, m;
    cin >> n >> m;
    if (n == 0 && m == 0){
      break;
    }
    vector<tuple<int, int, int>> E(m);
    for (int i = 0; i < m; i++){
      int s, t, c;
      cin >> s >> t >> c;
      s--;
      t--;
      E[i] = make_tuple(c, s, t);
    }
    sort(E.begin(), E.end());
    unionfind UF(n);
    int cnt = 0;
    for (int i = 0; i < m; i++){
      int c = get<0>(E[i]);
      int s = get<1>(E[i]);
      int t = get<2>(E[i]);
      if (!UF.same(s, t)){
        UF.unite(s, t);
        cnt++;
        if (cnt == n / 2){
          cout << c << endl;
          break;
        }
      }
    }
  }
}

```