#define  _CRT_SECURE_NO_WARNINGS
#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-11L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL

#define MOD 998244353LL
#define seg_size 262144 * 4LL
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

void init() {
	iostream::sync_with_stdio(false);
	cout << fixed << setprecision(20);
}

#define int ll

int union_tree[2000];
int union_find(int now) {
	if (union_tree[now] == now) return now;
	return union_tree[now] = union_find(union_tree[now]);
}
int union_merge(int a, int b) {
	a = union_find(a);
	b = union_find(b);
	if (a == b) return 0;
	union_tree[a] = b;
	return 1;
}

void solve(){
	while (true) {
		int n, m;
		cin >> n >> m;
		if (n == 0) return;
		vector<tuple<int, int, int>> inputs;
		REP(i, m) {
			int a, b, c;
			cin >> a >> b >> c;
			inputs.push_back(make_tuple(c, a, b));
		}
		sort(ALL(inputs));
		int target = n / 2;
		REP(i, n) {
			union_tree[i+1] = i + 1;
		}
		REP(i, inputs.size()) {
			target -= union_merge(get<1>(inputs[i]), get<2>(inputs[i]));
			if (target <= 0) {
				cout << get<0>(inputs[i]) << endl;
				break;
			}
		}
	}
}

#undef int
int main() {
	init();

	solve();
}

### 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
#define  _CRT_SECURE_NO_WARNINGS
#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-11L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL

#define MOD 998244353LL
#define seg_size 262144 * 4LL
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

void init() {
	iostream::sync_with_stdio(false);
	cout << fixed << setprecision(20);
}

#define int ll

int union_tree[2000];
int union_find(int now) {
	if (union_tree[now] == now) return now;
	return union_tree[now] = union_find(union_tree[now]);
}
int union_merge(int a, int b) {
	a = union_find(a);
	b = union_find(b);
	if (a == b) return 0;
	union_tree[a] = b;
	return 1;
}

void solve(){
	while (true) {
		int n, m;
		cin >> n >> m;
		if (n == 0) return;
		vector<tuple<int, int, int>> inputs;
		REP(i, m) {
			int a, b, c;
			cin >> a >> b >> c;
			inputs.push_back(make_tuple(c, a, b));
		}
		sort(ALL(inputs));
		int target = n / 2;
		REP(i, n) {
			union_tree[i+1] = i + 1;
		}
		REP(i, inputs.size()) {
			target -= union_merge(get<1>(inputs[i]), get<2>(inputs[i]));
			if (target <= 0) {
				cout << get<0>(inputs[i]) << endl;
				break;
			}
		}
	}
}

#undef int
int main() {
	init();

	solve();
}

```