#include <bits/stdc++.h>

#define rep(i,n) for(int i=0; i<n; i++)

using namespace std;

class UF {
	public:
	static const int ufmax=1052;
	int ufpar[ufmax],ufrank[ufmax],ufcnt[ufmax];
	UF() { init();}
	void init(){int i; rep(i,ufmax) { ufpar[i]=i; ufrank[i]=0; ufcnt[i]=1; } }
	int find(int x) {	return (ufpar[x]==x)?(x):(ufpar[x] = find(ufpar[x]));}
	int operator[](int x) {return find(x);}
	int count(int x) {return ufcnt[find(x)];}
	void unite(int x,int y) {
		x = find(x); y = find(y);
		if(x==y) return;
		if(ufrank[x]<ufrank[y]) ufpar[x]=y, ufcnt[y]+=ufcnt[x];
		else {ufpar[y]=x; ufcnt[x]+=ufcnt[y]; if(ufrank[x]==ufrank[y]) ufrank[x]++;}
	}
};

#define fi first
#define se second


int main(){
	int n,m;
	while(cin>>n>>m,n+m){
		using pii = pair<int,int>;
		using P = pair<int,pii>;
		vector<P> v(m);
		rep(i,m){
			cin>>v[i].se.fi>>v[i].se.se>>v[i].fi;
		}
		sort(v.begin(), v.end());
		UF uf;
		vector<int> w;
		for(auto x: v){
			int a=x.se.fi,b=x.se.se;
			a--,b--;
			if(uf[a] == uf[b]) continue;
			uf.unite(a,b);
			w.push_back(x.fi);
		}
		cout<<w[n/2-1]<<endl;
	}
}
### Prompt

In cpp, your task is to solve 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>

#define rep(i,n) for(int i=0; i<n; i++)

using namespace std;

class UF {
	public:
	static const int ufmax=1052;
	int ufpar[ufmax],ufrank[ufmax],ufcnt[ufmax];
	UF() { init();}
	void init(){int i; rep(i,ufmax) { ufpar[i]=i; ufrank[i]=0; ufcnt[i]=1; } }
	int find(int x) {	return (ufpar[x]==x)?(x):(ufpar[x] = find(ufpar[x]));}
	int operator[](int x) {return find(x);}
	int count(int x) {return ufcnt[find(x)];}
	void unite(int x,int y) {
		x = find(x); y = find(y);
		if(x==y) return;
		if(ufrank[x]<ufrank[y]) ufpar[x]=y, ufcnt[y]+=ufcnt[x];
		else {ufpar[y]=x; ufcnt[x]+=ufcnt[y]; if(ufrank[x]==ufrank[y]) ufrank[x]++;}
	}
};

#define fi first
#define se second


int main(){
	int n,m;
	while(cin>>n>>m,n+m){
		using pii = pair<int,int>;
		using P = pair<int,pii>;
		vector<P> v(m);
		rep(i,m){
			cin>>v[i].se.fi>>v[i].se.se>>v[i].fi;
		}
		sort(v.begin(), v.end());
		UF uf;
		vector<int> w;
		for(auto x: v){
			int a=x.se.fi,b=x.se.se;
			a--,b--;
			if(uf[a] == uf[b]) continue;
			uf.unite(a,b);
			w.push_back(x.fi);
		}
		cout<<w[n/2-1]<<endl;
	}
}
```