#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<int,P> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;

#define INF INT_MAX/3

#define MAX_N 1000


//// UnionFind
struct UnionFind{
	vector<int> v;
	UnionFind(int n) : v(n, -1) {}

	int find(int x) { return v[x] < 0 ? x : v[x] = find(v[x]); }
	bool unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y) return false;
		if (-v[x] < -v[y]) swap(x, y);
		v[x] += v[y]; v[y] = x;
		return true;
	}
	bool root(int x) { return v[x] < 0; }
	bool same(int x, int y) { return find(x) == find(y); }
	int size(int x) { return -v[find(x)]; }
};

int n,m;
int num;
PPI es[11111];

bool ok(ll x){
	set<int> st;
	UnionFind uf(n);
	rep(i,m){
		if(es[i].fi>x)break;
		uf.unite(es[i].se.fi,es[i].se.se);
	}
	rep(i,n){
		st.insert(uf.find(i));
	}
	return st.size()-1<n/2;
}

int main(){
	cin.sync_with_stdio(false);
	while(1){
		cin>>n>>m;
		if(n==0&&m==0)break;
		rep(i,m){
			cin>>es[i].se.fi>>es[i].se.se>>es[i].fi;
			es[i].se.fi--;
			es[i].se.se--;
		}
		sort(es,es+m);
		int lb=0,ub=1000;
		while(ub-lb>1){
			int mid=(lb+ub)/2;
			if(ok(mid))ub=mid;
			else lb=mid;
		}
		cout<<ub<<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;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<int,P> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;

#define INF INT_MAX/3

#define MAX_N 1000


//// UnionFind
struct UnionFind{
	vector<int> v;
	UnionFind(int n) : v(n, -1) {}

	int find(int x) { return v[x] < 0 ? x : v[x] = find(v[x]); }
	bool unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y) return false;
		if (-v[x] < -v[y]) swap(x, y);
		v[x] += v[y]; v[y] = x;
		return true;
	}
	bool root(int x) { return v[x] < 0; }
	bool same(int x, int y) { return find(x) == find(y); }
	int size(int x) { return -v[find(x)]; }
};

int n,m;
int num;
PPI es[11111];

bool ok(ll x){
	set<int> st;
	UnionFind uf(n);
	rep(i,m){
		if(es[i].fi>x)break;
		uf.unite(es[i].se.fi,es[i].se.se);
	}
	rep(i,n){
		st.insert(uf.find(i));
	}
	return st.size()-1<n/2;
}

int main(){
	cin.sync_with_stdio(false);
	while(1){
		cin>>n>>m;
		if(n==0&&m==0)break;
		rep(i,m){
			cin>>es[i].se.fi>>es[i].se.se>>es[i].fi;
			es[i].se.fi--;
			es[i].se.se--;
		}
		sort(es,es+m);
		int lb=0,ub=1000;
		while(ub-lb>1){
			int mid=(lb+ub)/2;
			if(ok(mid))ub=mid;
			else lb=mid;
		}
		cout<<ub<<endl;
	}
	return 0;
}
```