#include<iostream>
#include<queue>
#include<algorithm>
#define inf 1000000001
#define f first
#define s second
using namespace std;

int map[1000][1000];
int n,m,s,t,c;

int Prim()
{
  priority_queue<pair<int,int> > PQ;
  int d[1000],color[1000];
  for(int i=0;i<n;i++){
    d[i]=inf;
    color[i]=0;
  }
  d[0]=0;
  PQ.push(make_pair(0,0));
  color[0]=0;
  while(!PQ.empty()){
    pair<int,int> p=PQ.top();PQ.pop();
    int u=p.s;
    color[u]=2;

    if(d[u]<p.f*(-1))continue;

    for(int i=0;i<n;i++){
      if(color[i]==2 || map[u][i]==inf)continue;
      if(d[i]>map[u][i]){
	d[i]=map[u][i];
	PQ.push(make_pair(d[i]*(-1),i));
	color[i]=1;
      }
    }
  }
  sort(d,d+n);
  return d[n/2];
}  

int main()
{
  while(1){
    cin>>n>>m;
    if(n+m==0)break;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)map[i][j]=inf;
    for(int i=0;i<m;i++){
      cin>>s>>t>>c;
      map[s-1][t-1]=c;
      map[t-1][s-1]=c;
    }
    cout<<Prim()<<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<queue>
#include<algorithm>
#define inf 1000000001
#define f first
#define s second
using namespace std;

int map[1000][1000];
int n,m,s,t,c;

int Prim()
{
  priority_queue<pair<int,int> > PQ;
  int d[1000],color[1000];
  for(int i=0;i<n;i++){
    d[i]=inf;
    color[i]=0;
  }
  d[0]=0;
  PQ.push(make_pair(0,0));
  color[0]=0;
  while(!PQ.empty()){
    pair<int,int> p=PQ.top();PQ.pop();
    int u=p.s;
    color[u]=2;

    if(d[u]<p.f*(-1))continue;

    for(int i=0;i<n;i++){
      if(color[i]==2 || map[u][i]==inf)continue;
      if(d[i]>map[u][i]){
	d[i]=map[u][i];
	PQ.push(make_pair(d[i]*(-1),i));
	color[i]=1;
      }
    }
  }
  sort(d,d+n);
  return d[n/2];
}  

int main()
{
  while(1){
    cin>>n>>m;
    if(n+m==0)break;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)map[i][j]=inf;
    for(int i=0;i<m;i++){
      cin>>s>>t>>c;
      map[s-1][t-1]=c;
      map[t-1][s-1]=c;
    }
    cout<<Prim()<<endl;
  }
  return 0;
}
```