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

#define for_(i,a,b) for(int i=(a);i<(b);++i)

struct Edge {
	int u, v, c;
	bool operator < (const Edge& o) const { return c < o.c; }
};

class UnionFind {
public:
	vector< int > data;
	
	UnionFind(int n) : data(n, -1) {}
	
	void unionSet(int a, int b) {
		a = root(a); b = root(b);
		
		if (a != b) {
			if (data[a] < data[b]) swap(a, b);
			data[a] += data[b]; data[b] = a;
		}
	}
	
	bool sameSet(int a, int b) { return root(a) == root(b); }
	
	int root(int x) { return data[x] < 0 ? x : root(data[x]); }
};

int n, m;
vector< Edge > edges;

void solve() {
	sort(edges.begin(), edges.end());
	
	UnionFind uf(n);
	
	vector< int > cv;
	
	for_(i,0,m) {
		Edge e = edges[i];
		
		if (!uf.sameSet(e.u, e.v)) {
			uf.unionSet(e.u, e.v);
			cv.push_back(e.c);
		}
	}
	
	cout << cv[(n - 1) / 2] << endl;
}

int main() {
	while (cin >> n >> m, n) {
		edges.clear();
		
		for_(i,0,m) {
			int u, v, c;
			cin >> u >> v >> c;
			edges.push_back(Edge{--u, --v, c});
		}
		
		solve();
	}
}
### Prompt

Generate 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 for_(i,a,b) for(int i=(a);i<(b);++i)

struct Edge {
	int u, v, c;
	bool operator < (const Edge& o) const { return c < o.c; }
};

class UnionFind {
public:
	vector< int > data;
	
	UnionFind(int n) : data(n, -1) {}
	
	void unionSet(int a, int b) {
		a = root(a); b = root(b);
		
		if (a != b) {
			if (data[a] < data[b]) swap(a, b);
			data[a] += data[b]; data[b] = a;
		}
	}
	
	bool sameSet(int a, int b) { return root(a) == root(b); }
	
	int root(int x) { return data[x] < 0 ? x : root(data[x]); }
};

int n, m;
vector< Edge > edges;

void solve() {
	sort(edges.begin(), edges.end());
	
	UnionFind uf(n);
	
	vector< int > cv;
	
	for_(i,0,m) {
		Edge e = edges[i];
		
		if (!uf.sameSet(e.u, e.v)) {
			uf.unionSet(e.u, e.v);
			cv.push_back(e.c);
		}
	}
	
	cout << cv[(n - 1) / 2] << endl;
}

int main() {
	while (cin >> n >> m, n) {
		edges.clear();
		
		for_(i,0,m) {
			int u, v, c;
			cin >> u >> v >> c;
			edges.push_back(Edge{--u, --v, c});
		}
		
		solve();
	}
}
```