#include<iostream>
#include<vector>
#include<algorithm>
#include<deque>
#include<queue>
#include<cassert>
#include<climits>

#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define IINF (INT_MAX)
#define MAX 10010

using namespace std;

struct Data{
  int from,to,cost;
  Data(int from=-1,int to=-1,int cost=-1):from(from),to(to),cost(cost){}
};

bool cmp(const Data& p1,const Data& p2){ return p1.cost < p2.cost; }

int par[MAX+1];

void init(){ rep(i,MAX)par[i] = i; }

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

void unit(int x,int y){
  x = find(x);
  y = find(y);
  if( x != y)par[x] = y;
}

bool same(int x,int y){ return find(x) == find(y); }

int kruskal(vector<Data>& G){
  sort(G.begin(),G.end(),cmp);
  init();
  vector<int> vec;
  for(int i=0;i<G.size();i++){
    Data data = G[i];
    if(!same(data.from,data.to)){
      unit(data.from,data.to);
      vec.push_back(G[i].cost);
    }
  }

  sort(vec.begin(),vec.end());
  int len = vec.size();

  return vec[len/2];

}


int n,m;

int main(){

  while(cin >> n >> m,n|m){
    vector<Data> G;
    rep(i,m){
      int s,t,c;
      cin >> s >> t >> c;
      s--,t--;
      G.push_back(Data(s,t,c));
    }
    cout << kruskal(G) << endl;
  }

  return 0;
}
### Prompt

Your challenge is to write 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<iostream>
#include<vector>
#include<algorithm>
#include<deque>
#include<queue>
#include<cassert>
#include<climits>

#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define IINF (INT_MAX)
#define MAX 10010

using namespace std;

struct Data{
  int from,to,cost;
  Data(int from=-1,int to=-1,int cost=-1):from(from),to(to),cost(cost){}
};

bool cmp(const Data& p1,const Data& p2){ return p1.cost < p2.cost; }

int par[MAX+1];

void init(){ rep(i,MAX)par[i] = i; }

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

void unit(int x,int y){
  x = find(x);
  y = find(y);
  if( x != y)par[x] = y;
}

bool same(int x,int y){ return find(x) == find(y); }

int kruskal(vector<Data>& G){
  sort(G.begin(),G.end(),cmp);
  init();
  vector<int> vec;
  for(int i=0;i<G.size();i++){
    Data data = G[i];
    if(!same(data.from,data.to)){
      unit(data.from,data.to);
      vec.push_back(G[i].cost);
    }
  }

  sort(vec.begin(),vec.end());
  int len = vec.size();

  return vec[len/2];

}


int n,m;

int main(){

  while(cin >> n >> m,n|m){
    vector<Data> G;
    rep(i,m){
      int s,t,c;
      cin >> s >> t >> c;
      s--,t--;
      G.push_back(Data(s,t,c));
    }
    cout << kruskal(G) << endl;
  }

  return 0;
}
```