#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

class union_find {
public:
	union_find(int n)
	: par_(n, -1)
	{}
	void init(int n) {
		par_.assign(n, -1);
	}

	int root(int x) {
		return par_[x] < 0 ? x : par_[x] = root(par_[x]);
	}

	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if(x == y) {
			return false;
		} else {
			if(par_[x] < par_[y]) {
				par_[x] += par_[y];
				par_[y] = x;
			} else {
				par_[y] += par_[x];
				par_[x] = y;
			}
			return true;
		}
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}

	int size(int x) {
		return -par_[root(x)];
	}

private:
	std::vector<int> par_;
};

ll n,m;

ll solve() {
	union_find uf(n);
	ll cnct=n;
	vector<pair<ll,P>> edg;
	REP(i,m) {
		ll s,t,c;
		cin>>s>>t>>c;
		s--; t--;
		edg.pb(make_pair(c,P(s,t)));
	}
	sort(ALL(edg));
	REP(i,m) {
		if(!uf.same(edg[i].second.first,edg[i].second.second)) cnct--;
		if(cnct==n/2) return edg[i].first;
		uf.unite(edg[i].second.first,edg[i].second.second);
	}
	return 3939;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	while(1) {
		cin>>n>>m;
		if(n==0&&m==0) break;
		cout<<solve()<<endl;
	}
	return 0;
}


### 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<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

class union_find {
public:
	union_find(int n)
	: par_(n, -1)
	{}
	void init(int n) {
		par_.assign(n, -1);
	}

	int root(int x) {
		return par_[x] < 0 ? x : par_[x] = root(par_[x]);
	}

	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if(x == y) {
			return false;
		} else {
			if(par_[x] < par_[y]) {
				par_[x] += par_[y];
				par_[y] = x;
			} else {
				par_[y] += par_[x];
				par_[x] = y;
			}
			return true;
		}
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}

	int size(int x) {
		return -par_[root(x)];
	}

private:
	std::vector<int> par_;
};

ll n,m;

ll solve() {
	union_find uf(n);
	ll cnct=n;
	vector<pair<ll,P>> edg;
	REP(i,m) {
		ll s,t,c;
		cin>>s>>t>>c;
		s--; t--;
		edg.pb(make_pair(c,P(s,t)));
	}
	sort(ALL(edg));
	REP(i,m) {
		if(!uf.same(edg[i].second.first,edg[i].second.second)) cnct--;
		if(cnct==n/2) return edg[i].first;
		uf.unite(edg[i].second.first,edg[i].second.second);
	}
	return 3939;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	while(1) {
		cin>>n>>m;
		if(n==0&&m==0) break;
		cout<<solve()<<endl;
	}
	return 0;
}


```