#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
#define INF 100000000
#define EPS 1e-10
#define MOD 1000000007
using namespace std;
typedef pair<int,int> P;

struct UF{
	int par[1000];
	int rank[1000];

	void init(int n){
		rep(i,n){
			par[i] = i;
			rank[i] = 0;
		}
	}

	int find(int x){
		if(par[x] == x) return x;
		else return par[x] = find(par[x]);
	}

	void unite(int x, int y){
		x = find(x);
		y = find(y);
		if(x == y) return;
		
		if(rank[x] < rank[y]) par[x] = y;
		else{
			par[y] = x;
			if(rank[x] == rank[y]) rank[x]++;
		}
	}

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

int n, m;
vector<P> v[1000];
UF uf;

void solve(){
	rep(i,1000) v[i].clear();
	rep(i,m){
		int s, t, c;
		cin >> s >> t >> c;
		s--; t--;
		v[s].push_back(P(t,c));
	}
	int s = 0, e = 10000, mid;
	while(true){
		if(e-s == 1) break;
		mid = (s+e)/2;
		uf.init(n);
		rep(i,n){
			rep(j,v[i].size()){
				P p = v[i][j];
				if(p.second <= mid) uf.unite(i,p.first);
			}
		}
		int cnt = 0;
		rep(i,n) if(uf.par[i] == i) cnt++;
		if(cnt <= n/2) e = mid;
		else s = mid;
	}
	cout << e << endl;
}

int main(){
	while(cin >> n >> m){
		if(n == 0 && m == 0) break;
		solve();
	}
}
### Prompt

Please create a solution in CPP to 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++)
#define INF 100000000
#define EPS 1e-10
#define MOD 1000000007
using namespace std;
typedef pair<int,int> P;

struct UF{
	int par[1000];
	int rank[1000];

	void init(int n){
		rep(i,n){
			par[i] = i;
			rank[i] = 0;
		}
	}

	int find(int x){
		if(par[x] == x) return x;
		else return par[x] = find(par[x]);
	}

	void unite(int x, int y){
		x = find(x);
		y = find(y);
		if(x == y) return;
		
		if(rank[x] < rank[y]) par[x] = y;
		else{
			par[y] = x;
			if(rank[x] == rank[y]) rank[x]++;
		}
	}

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

int n, m;
vector<P> v[1000];
UF uf;

void solve(){
	rep(i,1000) v[i].clear();
	rep(i,m){
		int s, t, c;
		cin >> s >> t >> c;
		s--; t--;
		v[s].push_back(P(t,c));
	}
	int s = 0, e = 10000, mid;
	while(true){
		if(e-s == 1) break;
		mid = (s+e)/2;
		uf.init(n);
		rep(i,n){
			rep(j,v[i].size()){
				P p = v[i][j];
				if(p.second <= mid) uf.unite(i,p.first);
			}
		}
		int cnt = 0;
		rep(i,n) if(uf.par[i] == i) cnt++;
		if(cnt <= n/2) e = mid;
		else s = mid;
	}
	cout << e << endl;
}

int main(){
	while(cin >> n >> m){
		if(n == 0 && m == 0) break;
		solve();
	}
}
```