#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<29;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

class Union_Find_Tree{
	public:
	vi p,r,s;
	Union_Find_Tree(int n){
		p=r=vi(n);
		s=vi(n,1);
		for(int i=0;i<n;i++) p[i]=i;
	}
	int Par(int x){
		if(p[x]==x) return x;
		return p[x]=Par(p[x]);
	}
	int Size(int x){return s[Par(x)];}
	bool Unite(int x,int y){
		x=Par(x);
		y=Par(y);
		if(x==y) return 0;
		if(r[x]<r[y]){
			p[x]=y;
			s[y]+=s[x];
		}
		else{
			p[y]=x;
			s[x]+=s[y];
			if(r[x]==r[y]) r[x]++;
		}
		return 1;
	}
	bool Same(int x,int y){return Par(x)==Par(y);}
};

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	while(1){
		int n,m;
		cin>>n>>m;
		if(n==0) break;
		Union_Find_Tree uft(n);
		vip a;
		for(int i=0;i<m;i++){
			int u,v,w;
			cin>>u>>v>>w;
			u--;v--;
			a.push_back({w,{u,v}});
		}
		sort(a.begin(),a.end());
		vi b;
		for(int i=0;i<m;i++){
			int u=a[i].second.first,v=a[i].second.second;
			if(!uft.Same(u,v)){
				uft.Unite(u,v);
				b.push_back(a[i].first);
			}
		}
		cout<<b[n/2-1]<<endl;
	}
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<29;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

class Union_Find_Tree{
	public:
	vi p,r,s;
	Union_Find_Tree(int n){
		p=r=vi(n);
		s=vi(n,1);
		for(int i=0;i<n;i++) p[i]=i;
	}
	int Par(int x){
		if(p[x]==x) return x;
		return p[x]=Par(p[x]);
	}
	int Size(int x){return s[Par(x)];}
	bool Unite(int x,int y){
		x=Par(x);
		y=Par(y);
		if(x==y) return 0;
		if(r[x]<r[y]){
			p[x]=y;
			s[y]+=s[x];
		}
		else{
			p[y]=x;
			s[x]+=s[y];
			if(r[x]==r[y]) r[x]++;
		}
		return 1;
	}
	bool Same(int x,int y){return Par(x)==Par(y);}
};

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	while(1){
		int n,m;
		cin>>n>>m;
		if(n==0) break;
		Union_Find_Tree uft(n);
		vip a;
		for(int i=0;i<m;i++){
			int u,v,w;
			cin>>u>>v>>w;
			u--;v--;
			a.push_back({w,{u,v}});
		}
		sort(a.begin(),a.end());
		vi b;
		for(int i=0;i<m;i++){
			int u=a[i].second.first,v=a[i].second.second;
			if(!uft.Same(u,v)){
				uft.Unite(u,v);
				b.push_back(a[i].first);
			}
		}
		cout<<b[n/2-1]<<endl;
	}
}

```