#include<bits/stdc++.h>
using namespace std;
#define MAX 1005
#define INF (1<<21)
#define WHITE 0
#define GRAY  1
#define BLACK 2
int M[MAX][MAX];
int n,x;
int prim(){
  vector<int> v;
  int u,minv;
  int d[MAX],p[MAX],color[MAX];
  for(int i=0;i<n;i++){
    d[i]=INF;
    p[i]=-1;
    color[i]=WHITE;
  }
  d[0]=0;
  while(1){
    minv=INF;
    u=-1;
    for(int i=0;i<n;i++){
      if(minv>d[i]&&color[i]!=BLACK){
	u=i;
	minv=d[i];
      }
    }
    if(u==-1) break;
    color[u]=BLACK;
    for(int i=0;i<n;i++){
      if(color[i]!=BLACK&&M[u][i]!=INF){
	if(d[i]>M[u][i]){
	  d[i]=M[u][i];
	  p[i]=u;
	  color[i]=GRAY;
	}
      }
    }
  }
  for(int i=0;i<n;i++){
    if(~p[i]) v.push_back(M[i][p[i]]);
  }
  sort(v.begin(),v.end());
  if(v.size()%2) return v[v.size()/2];
  return (v[v.size()/2]+v[v.size()/2-1])/2;
}
int main(){
  while(cin>>n>>x,n||x){
    int s,t,c;
    for(int i=0;i<MAX;i++){
      for(int j=0;j<MAX;j++){
	M[i][j]=INF;
      }
    }
    for(int i=0;i<x;i++){
      cin>>s>>t>>c;
      M[s-1][t-1]=c;
      M[t-1][s-1]=c;
    }
    cout << prim() << endl;
  }
  return 0;
}
### Prompt

Generate 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>
using namespace std;
#define MAX 1005
#define INF (1<<21)
#define WHITE 0
#define GRAY  1
#define BLACK 2
int M[MAX][MAX];
int n,x;
int prim(){
  vector<int> v;
  int u,minv;
  int d[MAX],p[MAX],color[MAX];
  for(int i=0;i<n;i++){
    d[i]=INF;
    p[i]=-1;
    color[i]=WHITE;
  }
  d[0]=0;
  while(1){
    minv=INF;
    u=-1;
    for(int i=0;i<n;i++){
      if(minv>d[i]&&color[i]!=BLACK){
	u=i;
	minv=d[i];
      }
    }
    if(u==-1) break;
    color[u]=BLACK;
    for(int i=0;i<n;i++){
      if(color[i]!=BLACK&&M[u][i]!=INF){
	if(d[i]>M[u][i]){
	  d[i]=M[u][i];
	  p[i]=u;
	  color[i]=GRAY;
	}
      }
    }
  }
  for(int i=0;i<n;i++){
    if(~p[i]) v.push_back(M[i][p[i]]);
  }
  sort(v.begin(),v.end());
  if(v.size()%2) return v[v.size()/2];
  return (v[v.size()/2]+v[v.size()/2-1])/2;
}
int main(){
  while(cin>>n>>x,n||x){
    int s,t,c;
    for(int i=0;i<MAX;i++){
      for(int j=0;j<MAX;j++){
	M[i][j]=INF;
      }
    }
    for(int i=0;i<x;i++){
      cin>>s>>t>>c;
      M[s-1][t-1]=c;
      M[t-1][s-1]=c;
    }
    cout << prim() << endl;
  }
  return 0;
}
```