#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

#define lf double
#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

class UnionFind
{
private:
	vector<int> par;
	vector<int> rank; 
public:
	UnionFind(int n): par(n), rank(n)
	{
		for(int i=0; i<n; i++){
			par[i] = i;
			rank[i] = 0;
		}
	}
	int root(int x)
	{
		if (par[x] == x){
			return x;
		}else{
			return par[x] = root(par[x]);
		}
	}
	void unite(int x, int y)
	{
		int rx = root(x);
		int ry = root(y);
		if (rx == ry) return;
		if (rank[rx] > rank[ry]){
			par[ry] = rx;
		}else{
			par[rx] = ry;
			if (rank[rx] == rank[ry]) rank[ry]++;
		}
	}
	int same(int x, int y)
	{
		return root(x) == root(y);
	}
};

typedef struct {ll u, v, cost;} edge;

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n, m;
	while(cin>>n>>m, n>0){
		vector<edge> es(m);
		rep(i, m){
			ll s, t, c;
			cin>>s>>t>>c;
			es[i].u = s-1;
			es[i].v = t-1;
			es[i].cost = c;
		}
		UnionFind uf(n);
		vll ans;
		sort(es.begin(), es.end(),
			[](edge &x, edge &y){return x.cost < y.cost;});
		rep(i, m){
			edge e = es[i];
			if (!uf.same(e.u, e.v)){
				uf.unite(e.u, e.v);
				ans.pb(e.cost);
			}
		}
		cout << ans[ans.size()/2] << "\n";
	}
	return 0;
}

### 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 <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

#define lf double
#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

class UnionFind
{
private:
	vector<int> par;
	vector<int> rank; 
public:
	UnionFind(int n): par(n), rank(n)
	{
		for(int i=0; i<n; i++){
			par[i] = i;
			rank[i] = 0;
		}
	}
	int root(int x)
	{
		if (par[x] == x){
			return x;
		}else{
			return par[x] = root(par[x]);
		}
	}
	void unite(int x, int y)
	{
		int rx = root(x);
		int ry = root(y);
		if (rx == ry) return;
		if (rank[rx] > rank[ry]){
			par[ry] = rx;
		}else{
			par[rx] = ry;
			if (rank[rx] == rank[ry]) rank[ry]++;
		}
	}
	int same(int x, int y)
	{
		return root(x) == root(y);
	}
};

typedef struct {ll u, v, cost;} edge;

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n, m;
	while(cin>>n>>m, n>0){
		vector<edge> es(m);
		rep(i, m){
			ll s, t, c;
			cin>>s>>t>>c;
			es[i].u = s-1;
			es[i].v = t-1;
			es[i].cost = c;
		}
		UnionFind uf(n);
		vll ans;
		sort(es.begin(), es.end(),
			[](edge &x, edge &y){return x.cost < y.cost;});
		rep(i, m){
			edge e = es[i];
			if (!uf.same(e.u, e.v)){
				uf.unite(e.u, e.v);
				ans.pb(e.cost);
			}
		}
		cout << ans[ans.size()/2] << "\n";
	}
	return 0;
}

```