#include <algorithm>
#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;

class union_find {
private:
	vector<int> uf;
public:
	union_find(int n):uf(n, -1) {}

	int find(int x) {
		return uf[x] < 0 ? x : uf[x] = find(uf[x]);
	}

	bool unite(int a, int b) {
		a = find(a);
		b = find(b);

		if(a == b) return false;

		if(uf[a] > uf[b]) {
			swap(a, b);
		}
		else if(uf[a] == uf[b]) {
			--uf[a];
		}

		uf[b] = a;
		return true;
	}
};

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

int calc(int n, vector<edge> &edges) {
	sort(edges.begin(), edges.end());

	int cnt = 0;
	union_find uf(n);
	for(const auto &e : edges) {
		if(uf.unite(e.u, e.v) && ++cnt == n / 2) return e.cost;
	}
}

int main() {
	for(int n, m; cin >> n >> m && n;) {
		vector<edge> edges;
		edges.reserve(m);
		for(int i = 0; i < m; ++i) {
			int s, t, c;
			cin >> s >> t >> c;
			edges.emplace_back(s - 1, t - 1, c);
		}
		cout << calc(n, edges) << endl;
	}

	return EXIT_SUCCESS;
}
### 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 <algorithm>
#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;

class union_find {
private:
	vector<int> uf;
public:
	union_find(int n):uf(n, -1) {}

	int find(int x) {
		return uf[x] < 0 ? x : uf[x] = find(uf[x]);
	}

	bool unite(int a, int b) {
		a = find(a);
		b = find(b);

		if(a == b) return false;

		if(uf[a] > uf[b]) {
			swap(a, b);
		}
		else if(uf[a] == uf[b]) {
			--uf[a];
		}

		uf[b] = a;
		return true;
	}
};

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

int calc(int n, vector<edge> &edges) {
	sort(edges.begin(), edges.end());

	int cnt = 0;
	union_find uf(n);
	for(const auto &e : edges) {
		if(uf.unite(e.u, e.v) && ++cnt == n / 2) return e.cost;
	}
}

int main() {
	for(int n, m; cin >> n >> m && n;) {
		vector<edge> edges;
		edges.reserve(m);
		for(int i = 0; i < m; ++i) {
			int s, t, c;
			cin >> s >> t >> c;
			edges.emplace_back(s - 1, t - 1, c);
		}
		cout << calc(n, edges) << endl;
	}

	return EXIT_SUCCESS;
}
```