#ifndef ___Rank_Union_Find
#define ___Rank_Union_Find

#include <vector>

// ------ Class ------ //
class UnionFind {
private:
	unsigned size_; std::vector<unsigned> par, rank;
public:
	UnionFind() : size_(0), par(std::vector<unsigned>()), rank(std::vector<unsigned>()) {};
	UnionFind(unsigned size__) : size_(size__) {
		par.resize(size_); rank.resize(size_);
		for (unsigned i = 0; i < size_; i++) par[i] = i, rank[i] = 0;
	}
	unsigned size() { return size_; }
	unsigned root(unsigned x) { return par[x] == x ? x : par[x] = root(par[x]); }
	bool same(unsigned x, unsigned y) { return root(x) == root(y); }
	void unite(unsigned x, unsigned y) {
		x = root(x), y = root(y);
		if (x == y) return;
		if (rank[x] < rank[y]) par[x] = y;
		else if (rank[x] == rank[y]) par[y] = x, rank[x]++;
		else par[y] = x;
	}
	bool operator==(const UnionFind &u) { return par == u.par; }
	bool operator!=(const UnionFind &u) { return par != u.par; }
};

#endif

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
struct edge { int s, t, c; };
bool operator<(const edge& e1, const edge& e2) { return e1.c < e2.c; }
int n, m;
int main() {
	while (scanf("%d%d", &n, &m), n | m) {
		vector<edge> v(m);
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &v[i].s, &v[i].t, &v[i].c);
			v[i].s--, v[i].t--;
		}
		sort(v.begin(), v.end());
		int cnt = 0;
		UnionFind uf(n);
		for (int i = 0; i < n; i++) {
			if (!uf.same(v[i].s, v[i].t)) {
				uf.unite(v[i].s, v[i].t);
				cnt++;
			}
			if (cnt == n / 2) {
				printf("%d\n", v[i].c);
				break;
			}
		}
	}
	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
#ifndef ___Rank_Union_Find
#define ___Rank_Union_Find

#include <vector>

// ------ Class ------ //
class UnionFind {
private:
	unsigned size_; std::vector<unsigned> par, rank;
public:
	UnionFind() : size_(0), par(std::vector<unsigned>()), rank(std::vector<unsigned>()) {};
	UnionFind(unsigned size__) : size_(size__) {
		par.resize(size_); rank.resize(size_);
		for (unsigned i = 0; i < size_; i++) par[i] = i, rank[i] = 0;
	}
	unsigned size() { return size_; }
	unsigned root(unsigned x) { return par[x] == x ? x : par[x] = root(par[x]); }
	bool same(unsigned x, unsigned y) { return root(x) == root(y); }
	void unite(unsigned x, unsigned y) {
		x = root(x), y = root(y);
		if (x == y) return;
		if (rank[x] < rank[y]) par[x] = y;
		else if (rank[x] == rank[y]) par[y] = x, rank[x]++;
		else par[y] = x;
	}
	bool operator==(const UnionFind &u) { return par == u.par; }
	bool operator!=(const UnionFind &u) { return par != u.par; }
};

#endif

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
struct edge { int s, t, c; };
bool operator<(const edge& e1, const edge& e2) { return e1.c < e2.c; }
int n, m;
int main() {
	while (scanf("%d%d", &n, &m), n | m) {
		vector<edge> v(m);
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &v[i].s, &v[i].t, &v[i].c);
			v[i].s--, v[i].t--;
		}
		sort(v.begin(), v.end());
		int cnt = 0;
		UnionFind uf(n);
		for (int i = 0; i < n; i++) {
			if (!uf.same(v[i].s, v[i].t)) {
				uf.unite(v[i].s, v[i].t);
				cnt++;
			}
			if (cnt == n / 2) {
				printf("%d\n", v[i].c);
				break;
			}
		}
	}
	return 0;
}
```