#include <bits/stdc++.h>
typedef long long LL;
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

using namespace std;

struct Unionfind {
  vector<int> size;
  vector<int> par;

  Unionfind( int n ) :  size(n, 1), par(n) {
    iota( par.begin() , par.end() , 0 );
  }

  int find( int x ) {
    if ( par[x] == x ) return x;
    return par[x] = find( par[x] );
  }

  bool unite( int x , int y ) {
    x = find(x);
    y = find(y);
    if ( x == y ) return false;

    if ( size[y] < size[x] ) swap( x , y );
    par[x] = y;
    size[y] += size[x];

    return true;
  }

  bool same( int x , int y ) {
    return find(x) == find(y);
  }
};
int main(void)
{
  for(;;){
    int n,m;
    cin >> n >> m;
    if(!n&&!m) return 0;
    Unionfind t(n+1);
    vector<tuple<int,int,int> > e(m);
    REP(i,m){
      int a,b,c;
      cin >> a >> b >> c;
      e[i]=make_tuple(c,a,b);
    }
    SORT(e);
    int tmp=0,answer;
    REP(i,m){
      int a,b,c;
      tie(c,a,b)=e[i];
      if(t.unite(a,b)){
	++tmp;
	if(tmp*2==n) answer=c;
      }
    }
    cout << answer << endl;
  }
}
### Prompt

Your task is to create a Cpp solution to 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
#include <bits/stdc++.h>
typedef long long LL;
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

using namespace std;

struct Unionfind {
  vector<int> size;
  vector<int> par;

  Unionfind( int n ) :  size(n, 1), par(n) {
    iota( par.begin() , par.end() , 0 );
  }

  int find( int x ) {
    if ( par[x] == x ) return x;
    return par[x] = find( par[x] );
  }

  bool unite( int x , int y ) {
    x = find(x);
    y = find(y);
    if ( x == y ) return false;

    if ( size[y] < size[x] ) swap( x , y );
    par[x] = y;
    size[y] += size[x];

    return true;
  }

  bool same( int x , int y ) {
    return find(x) == find(y);
  }
};
int main(void)
{
  for(;;){
    int n,m;
    cin >> n >> m;
    if(!n&&!m) return 0;
    Unionfind t(n+1);
    vector<tuple<int,int,int> > e(m);
    REP(i,m){
      int a,b,c;
      cin >> a >> b >> c;
      e[i]=make_tuple(c,a,b);
    }
    SORT(e);
    int tmp=0,answer;
    REP(i,m){
      int a,b,c;
      tie(c,a,b)=e[i];
      if(t.unite(a,b)){
	++tmp;
	if(tmp*2==n) answer=c;
      }
    }
    cout << answer << endl;
  }
}
```