#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <cstring>
#include <functional>
#include <cmath>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep1(i,n) for(int i=1;i<=(n);++i)
#define all(c) (c).begin(),(c).end()
#define fs first
#define sc second
#define pb push_back
#define show(x) cout << #x << " " << x << endl
struct edge{int from,to,cost;};
vector<edge> es;
int par[1000];
void init(int n){
	rep(i,n) par[i]=i;
}
int find(int x){
	if(x==par[x]) return x;
	return par[x]=find(par[x]);
}
bool same(int x,int y){
	return find(x)==find(y);
}
void unite(int x,int y){
	x=find(x),y=find(y);
	if(x!=y) par[x]=y;
}
bool comp(const edge &x,const edge &y){
	if(x.cost!=y.cost) return x.cost<y.cost;
	if(x.from!=y.from) return x.from<y.from;
	return x.to<y.to;
}
int main(){
	while(true){
		int n,m;
		cin>>n>>m;
		if(n==0) break;
		init(n);
		es.clear();
		rep(i,m){
			int a,b,c;
			cin>>a>>b>>c;
			es.pb(edge{a-1,b-1,c});
		}
		sort(all(es),comp);
		int cnt=0;
		for(auto e:es){
			if(same(e.from,e.to)) continue;
			unite(e.from,e.to);
			cnt++;
			if(cnt==n/2){
				cout<<e.cost<<endl;
				break;
			}
		}
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <cstring>
#include <functional>
#include <cmath>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep1(i,n) for(int i=1;i<=(n);++i)
#define all(c) (c).begin(),(c).end()
#define fs first
#define sc second
#define pb push_back
#define show(x) cout << #x << " " << x << endl
struct edge{int from,to,cost;};
vector<edge> es;
int par[1000];
void init(int n){
	rep(i,n) par[i]=i;
}
int find(int x){
	if(x==par[x]) return x;
	return par[x]=find(par[x]);
}
bool same(int x,int y){
	return find(x)==find(y);
}
void unite(int x,int y){
	x=find(x),y=find(y);
	if(x!=y) par[x]=y;
}
bool comp(const edge &x,const edge &y){
	if(x.cost!=y.cost) return x.cost<y.cost;
	if(x.from!=y.from) return x.from<y.from;
	return x.to<y.to;
}
int main(){
	while(true){
		int n,m;
		cin>>n>>m;
		if(n==0) break;
		init(n);
		es.clear();
		rep(i,m){
			int a,b,c;
			cin>>a>>b>>c;
			es.pb(edge{a-1,b-1,c});
		}
		sort(all(es),comp);
		int cnt=0;
		for(auto e:es){
			if(same(e.from,e.to)) continue;
			unite(e.from,e.to);
			cnt++;
			if(cnt==n/2){
				cout<<e.cost<<endl;
				break;
			}
		}
	}
}
```