#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)

template<int um> class UF {
public:
	vector<int> par, rank;
	UF() { rank = vector<int>(um, 0); for (int i = 0; i<um; i++) par.push_back(i); }
	int operator[](int x) { return (par[x] == x) ? (x) : (par[x] = operator[](par[x])); }
	int operator()(int x, int y) {
		if ((x = operator[](x)) == (y = operator[](y))) return x;
		if (rank[x]>rank[y]) return par[x] = y;
		rank[x] += rank[x] == rank[y]; return par[y] = x;
	}
};

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	while (1)
	{
		int n, m; cin >> n >> m;
		if (n == 0) return 0;
		
		vector< pair<int, int> > E;
		rep(i, 0, m)
		{
			int s, t, c;
			cin >> s >> t >> c;
			s--;
			t--;
			E.push_back(make_pair(c, s * 1000 + t));
		}

		sort(E.begin(), E.end());

		UF<1010> uf;
		vector<int> costs;
		int cnt = 0;

		for (auto p : E)
		{
			int s = p.second / 1000;
			int t = p.second % 1000;
			int c = p.first;

			if (uf[s] == uf[t]) continue;
			
			uf(s, t);
			costs.push_back(c);
			cnt++;

			if (cnt == n - 1) break;
		}

		sort(costs.begin(), costs.end());
		cout << costs[n / 2 - 1] << endl;
	}
}
### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)

template<int um> class UF {
public:
	vector<int> par, rank;
	UF() { rank = vector<int>(um, 0); for (int i = 0; i<um; i++) par.push_back(i); }
	int operator[](int x) { return (par[x] == x) ? (x) : (par[x] = operator[](par[x])); }
	int operator()(int x, int y) {
		if ((x = operator[](x)) == (y = operator[](y))) return x;
		if (rank[x]>rank[y]) return par[x] = y;
		rank[x] += rank[x] == rank[y]; return par[y] = x;
	}
};

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	while (1)
	{
		int n, m; cin >> n >> m;
		if (n == 0) return 0;
		
		vector< pair<int, int> > E;
		rep(i, 0, m)
		{
			int s, t, c;
			cin >> s >> t >> c;
			s--;
			t--;
			E.push_back(make_pair(c, s * 1000 + t));
		}

		sort(E.begin(), E.end());

		UF<1010> uf;
		vector<int> costs;
		int cnt = 0;

		for (auto p : E)
		{
			int s = p.second / 1000;
			int t = p.second % 1000;
			int c = p.first;

			if (uf[s] == uf[t]) continue;
			
			uf(s, t);
			costs.push_back(c);
			cnt++;

			if (cnt == n - 1) break;
		}

		sort(costs.begin(), costs.end());
		cout << costs[n / 2 - 1] << endl;
	}
}
```