#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#include<vector>
struct UF{
	int n;
	vector<int>parent,rank;
	UF(int n_=0):n(n_),parent(n_),rank(n_,1)
	{
		for(int i=0;i<n_;i++)parent[i]=i;
	}
	int find(int a)
	{
		return parent[a]!=a?parent[a]=find(parent[a]):a;
	}
	bool same(int a,int b)
	{
		return find(a)==find(b);
	}
	bool unite(int a,int b)
	{
		a=find(a),b=find(b);
		if(a==b)return false;
		if(rank[a]<rank[b])
		{
			parent[a]=b;
			rank[b]+=rank[a];
		}
		else
		{
			parent[b]=a;
			rank[a]+=rank[b];
		}
		return true;
	}
};
int n,m;
int main()
{
	while(cin>>n>>m,n)
	{
		vector<pair<int,pair<int,int> > >E;
		for(int i=0;i<m;i++)
		{
			int s,t,c;cin>>s>>t>>c;
			E.push_back(make_pair(c,make_pair(s,t)));
		}
		sort(E.begin(),E.end());
		UF uf(n+1);
		int ans,cnt=0;
		for(int i=0;i<m;i++)
		{
			if(uf.unite(E[i].second.first,E[i].second.second))
			{
				cnt++;
				if(cnt==n/2)ans=E[i].first;
			}
		}
		cout<<ans<<endl;
	}
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
#include<vector>
struct UF{
	int n;
	vector<int>parent,rank;
	UF(int n_=0):n(n_),parent(n_),rank(n_,1)
	{
		for(int i=0;i<n_;i++)parent[i]=i;
	}
	int find(int a)
	{
		return parent[a]!=a?parent[a]=find(parent[a]):a;
	}
	bool same(int a,int b)
	{
		return find(a)==find(b);
	}
	bool unite(int a,int b)
	{
		a=find(a),b=find(b);
		if(a==b)return false;
		if(rank[a]<rank[b])
		{
			parent[a]=b;
			rank[b]+=rank[a];
		}
		else
		{
			parent[b]=a;
			rank[a]+=rank[b];
		}
		return true;
	}
};
int n,m;
int main()
{
	while(cin>>n>>m,n)
	{
		vector<pair<int,pair<int,int> > >E;
		for(int i=0;i<m;i++)
		{
			int s,t,c;cin>>s>>t>>c;
			E.push_back(make_pair(c,make_pair(s,t)));
		}
		sort(E.begin(),E.end());
		UF uf(n+1);
		int ans,cnt=0;
		for(int i=0;i<m;i++)
		{
			if(uf.unite(E[i].second.first,E[i].second.second))
			{
				cnt++;
				if(cnt==n/2)ans=E[i].first;
			}
		}
		cout<<ans<<endl;
	}
}

```