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

class union_find {
	vector<int> par;
	int find(int a) {
		return par[a] < 0 ? a : par[a] = find(par[a]);
	}
public:
	union_find(int n) : par(n, -1) {}
	bool same(int a, int b) {
		return find(a) == find(b);
	}
	void unite(int a, int b) {
		a = find(a);
		b = find(b);
		if (a == b) return;
		if (par[a] < par[b]) {
			par[a] += par[b];
			par[b] = a;
		}
		else {
			par[b] += par[a];
			par[a] = b;
		}
	}
};

struct edge {
	int from, to;
	int cost;
	bool operator<(const edge& e) const {
		return cost < e.cost;
	}
};

int main()
{
	int n, m;
	while (cin >> n >> m, n | m) {
		vector<edge> es;
		for (int i = 0; i < m; i++) {
			int s, t, c;
			cin >> s >> t >> c; --s, --t;
			es.push_back((edge) { s, t, c });
		}
		sort(es.begin(), es.end());
		vector<int> v;
		union_find uf(n);
		for (int i = 0; i < m; i++) {
			if (!uf.same(es[i].from, es[i].to)) {
				uf.unite(es[i].from, es[i].to);
				v.push_back(es[i].cost);
			}
		}
		cout << v[n / 2 - 1] << endl;
	}
	return 0;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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;

class union_find {
	vector<int> par;
	int find(int a) {
		return par[a] < 0 ? a : par[a] = find(par[a]);
	}
public:
	union_find(int n) : par(n, -1) {}
	bool same(int a, int b) {
		return find(a) == find(b);
	}
	void unite(int a, int b) {
		a = find(a);
		b = find(b);
		if (a == b) return;
		if (par[a] < par[b]) {
			par[a] += par[b];
			par[b] = a;
		}
		else {
			par[b] += par[a];
			par[a] = b;
		}
	}
};

struct edge {
	int from, to;
	int cost;
	bool operator<(const edge& e) const {
		return cost < e.cost;
	}
};

int main()
{
	int n, m;
	while (cin >> n >> m, n | m) {
		vector<edge> es;
		for (int i = 0; i < m; i++) {
			int s, t, c;
			cin >> s >> t >> c; --s, --t;
			es.push_back((edge) { s, t, c });
		}
		sort(es.begin(), es.end());
		vector<int> v;
		union_find uf(n);
		for (int i = 0; i < m; i++) {
			if (!uf.same(es[i].from, es[i].to)) {
				uf.unite(es[i].from, es[i].to);
				v.push_back(es[i].cost);
			}
		}
		cout << v[n / 2 - 1] << endl;
	}
	return 0;
}


```