#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#define fs first
#define sc second
#define pb push_back
using namespace std;
 
typedef pair<int,int> P;
struct edge{int u,v,cost;};
bool comp(const edge& e1, const edge& e2){return e1.cost<e2.cost;}
 
int n,m;
int s,t,c;
edge e[10100];
vector<int> sp;
vector<P> g[1010];
bool use[1010],h[10100];
 
int dfs(int pos){
  if(use[pos])return 0;
  use[pos] = true;
  int res = 1;
  for(int i=0;i<g[pos].size();i++){
    if(h[g[pos][i].sc])continue;
    res += dfs(g[pos][i].fs);
  }
  return res;
}
 
int main(){
  while(cin >> n >> m,n||m){
    for(int i=0;i<m;i++){
      sp.clear();
      cin >> s >> t >> c;
      s--; t--;
      e[i].u = s;
      e[i].v = t;
      e[i].cost = c;
    }
 
    sort(e,e+m,comp);
    sp.clear();    
 
    for(int i=0;i<n;i++)g[i].clear();
    
    for(int i=0;i<m;i++){
      g[e[i].u].pb(P(e[i].v,i));
      g[e[i].v].pb(P(e[i].u,i));
      h[i] = false;
    }
 
    for(int i=m-1;i>=0;i--){
      for(int j=0;j<n;j++)use[j] = false;
      h[i] = true;
      if(dfs(0) != n){
	h[i] = false;
	sp.pb(e[i].cost);
      }
    }
        
    sort(sp.begin(),sp.end());
    cout << sp[n/2-1] << endl;
  }
}
### 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<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#define fs first
#define sc second
#define pb push_back
using namespace std;
 
typedef pair<int,int> P;
struct edge{int u,v,cost;};
bool comp(const edge& e1, const edge& e2){return e1.cost<e2.cost;}
 
int n,m;
int s,t,c;
edge e[10100];
vector<int> sp;
vector<P> g[1010];
bool use[1010],h[10100];
 
int dfs(int pos){
  if(use[pos])return 0;
  use[pos] = true;
  int res = 1;
  for(int i=0;i<g[pos].size();i++){
    if(h[g[pos][i].sc])continue;
    res += dfs(g[pos][i].fs);
  }
  return res;
}
 
int main(){
  while(cin >> n >> m,n||m){
    for(int i=0;i<m;i++){
      sp.clear();
      cin >> s >> t >> c;
      s--; t--;
      e[i].u = s;
      e[i].v = t;
      e[i].cost = c;
    }
 
    sort(e,e+m,comp);
    sp.clear();    
 
    for(int i=0;i<n;i++)g[i].clear();
    
    for(int i=0;i<m;i++){
      g[e[i].u].pb(P(e[i].v,i));
      g[e[i].v].pb(P(e[i].u,i));
      h[i] = false;
    }
 
    for(int i=m-1;i>=0;i--){
      for(int j=0;j<n;j++)use[j] = false;
      h[i] = true;
      if(dfs(0) != n){
	h[i] = false;
	sp.pb(e[i].cost);
      }
    }
        
    sort(sp.begin(),sp.end());
    cout << sp[n/2-1] << endl;
  }
}
```