#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
int par[100000];
int rank1[100000];
void init(int n) {
	for (int i = 0; i < n; i++) {
		par[i] = i;
		rank1[i] = 0;
	}
}
int find1(int x) {
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find1(par[x]);
	}
}
void unite1(int x, int y) {
	x = find1(x);
	y = find1(y);
	if (x == y)return;
	if (rank1[x] < rank1[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (rank1[x] == rank1[y])rank1[x]++;
	}
}
bool same(int x, int y) {
	return find1(x) == find1(y);
}
struct edge { int u, v, cost; };
bool comp(const edge&e1, const edge&e2) {
	return e1.cost < e2.cost;
}
edge es[100000];
int V, E; vector<int>v;
void kruskal() {
	sort(es, es + E, comp);
	init(V); v.clear();
	for (int i = 0; i < E; i++) {
		edge e = es[i];
		if (!same(e.u, e.v)) {
			unite1(e.u, e.v);
			v.push_back(e.cost);
		}
	}
}
int main() {
	while (cin >> V >> E, V) {
		int x, y, z;
		rep(i, E) {
			cin >> x >> y >> z; x--; y--;
			es[i] = { x,y,z };
		}
		kruskal();
		cout << v[V/ 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<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
int par[100000];
int rank1[100000];
void init(int n) {
	for (int i = 0; i < n; i++) {
		par[i] = i;
		rank1[i] = 0;
	}
}
int find1(int x) {
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find1(par[x]);
	}
}
void unite1(int x, int y) {
	x = find1(x);
	y = find1(y);
	if (x == y)return;
	if (rank1[x] < rank1[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (rank1[x] == rank1[y])rank1[x]++;
	}
}
bool same(int x, int y) {
	return find1(x) == find1(y);
}
struct edge { int u, v, cost; };
bool comp(const edge&e1, const edge&e2) {
	return e1.cost < e2.cost;
}
edge es[100000];
int V, E; vector<int>v;
void kruskal() {
	sort(es, es + E, comp);
	init(V); v.clear();
	for (int i = 0; i < E; i++) {
		edge e = es[i];
		if (!same(e.u, e.v)) {
			unite1(e.u, e.v);
			v.push_back(e.cost);
		}
	}
}
int main() {
	while (cin >> V >> E, V) {
		int x, y, z;
		rep(i, E) {
			cin >> x >> y >> z; x--; y--;
			es[i] = { x,y,z };
		}
		kruskal();
		cout << v[V/ 2-1] << endl;
	}
	return 0;
}


```