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

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

class QuickFind {
private:
	int n;
	vector< vector< int > > group;
	vector< int > mapping;
	
public:
	QuickFind(int __n) : n(__n), group(__n, vector< int >()), mapping(__n) {
		for_(i,0,n) {
			group[i].push_back(i);
			mapping[i] = i;
		}
	}
	
	void merge(int x, int y) {
		int xi = mapping[x], yi = mapping[y];
		if (group[xi].size() > group[yi].size()) swap(xi, yi);
		for (int i : group[xi]) {
			group[yi].push_back(i);
			mapping[i] = yi;
		}
		group[xi] = vector< int >();
	}
	
	bool same(int x, int y) { return mapping[x] == mapping[y]; }
};

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

int main() {
	int n, m;
	while (cin >> n >> m, n > 0) {
		vector< Edge > edge;
		for_(i,0,m) {
			int u, v, c;
			cin >> u >> v >> c;
			--u; --v;
			edge.push_back(Edge{u, v, c});
		}
		
		sort(edge.begin(), edge.end());
		
		QuickFind qf(n);
		
		int cnt = n / 2;
		for_(i,0,m) {
			Edge e = edge[i];
			if (qf.same(e.u, e.v)) continue;
			qf.merge(e.u, e.v);
			--cnt;
			if (cnt == 0) {
				cout << e.c << endl;
				break;
			}
		}
	}
}
### Prompt

Please provide a Cpp coded solution 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>
using namespace std;

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

class QuickFind {
private:
	int n;
	vector< vector< int > > group;
	vector< int > mapping;
	
public:
	QuickFind(int __n) : n(__n), group(__n, vector< int >()), mapping(__n) {
		for_(i,0,n) {
			group[i].push_back(i);
			mapping[i] = i;
		}
	}
	
	void merge(int x, int y) {
		int xi = mapping[x], yi = mapping[y];
		if (group[xi].size() > group[yi].size()) swap(xi, yi);
		for (int i : group[xi]) {
			group[yi].push_back(i);
			mapping[i] = yi;
		}
		group[xi] = vector< int >();
	}
	
	bool same(int x, int y) { return mapping[x] == mapping[y]; }
};

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

int main() {
	int n, m;
	while (cin >> n >> m, n > 0) {
		vector< Edge > edge;
		for_(i,0,m) {
			int u, v, c;
			cin >> u >> v >> c;
			--u; --v;
			edge.push_back(Edge{u, v, c});
		}
		
		sort(edge.begin(), edge.end());
		
		QuickFind qf(n);
		
		int cnt = n / 2;
		for_(i,0,m) {
			Edge e = edge[i];
			if (qf.same(e.u, e.v)) continue;
			qf.merge(e.u, e.v);
			--cnt;
			if (cnt == 0) {
				cout << e.c << endl;
				break;
			}
		}
	}
}
```