#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
#ifdef int
const ll INF = (1LL<<60);
#else
const int INF = (1LL<<30);
#endif
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

class UnionFind {
public:
  const static int MAX_N = 100010;
  int par[MAX_N];
  int rank[MAX_N];
  int s[MAX_N];
  bool used[MAX_N];
  UnionFind() { init(); }
  UnionFind(int n) { init(n); }
  void init() { for(int i=0; i<MAX_N; ++i) par[i] = i, rank[i] = 0, s[i] = 1; }
  void init(int n) { for(int i=0; i<n; ++i) par[i] = i, rank[i] = 0, s[i] = 1; }
  int find(int x) {
    if(par[x] == x) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return;
    if(rank[x] < rank[y]) {
      par[x] = y; s[y] = s[x] + s[y];
    } else {
      par[y] = x; s[x] = s[x] + s[y];
      if( rank[x] == rank[y] ) rank[x]++;
    }
  }
  int size(int x) { return s[find(x)];}
  bool same(int x, int y) { return find(x) == find(y);}
  int group(int n) {
    REP(i, n) used[find(i)] = true;
    int ret = 0;
    REP(i, n) if(used[i]) ret++;
    return ret;
  }
};
UnionFind uf;

VI a[10010];
signed main(void)
{
  while(true) {
    int n, m;
    cin >> n >> m;
    if(!n) break;
    REP(i, m) {
      int x, y, z;
      cin >> x >> y >> z;
      x--, y--;
      a[i] = {z, x, y};
    }
    sort(a, a+m);
    if(n == 2) {
      cout << a[0][0] << endl;
      continue;
    }

    uf.init(n);
    VI v;
    REP(i, m) {
      if(!uf.same(a[i][1], a[i][2])) {
        v.PB(a[i][0]);
        uf.unite(a[i][1], a[i][2]);
      }
    }
    sort(ALL(v));
    cout << v[n/2-1] << endl;
  }

  return 0;
}
### Prompt

Your task is to create 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
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
#ifdef int
const ll INF = (1LL<<60);
#else
const int INF = (1LL<<30);
#endif
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

class UnionFind {
public:
  const static int MAX_N = 100010;
  int par[MAX_N];
  int rank[MAX_N];
  int s[MAX_N];
  bool used[MAX_N];
  UnionFind() { init(); }
  UnionFind(int n) { init(n); }
  void init() { for(int i=0; i<MAX_N; ++i) par[i] = i, rank[i] = 0, s[i] = 1; }
  void init(int n) { for(int i=0; i<n; ++i) par[i] = i, rank[i] = 0, s[i] = 1; }
  int find(int x) {
    if(par[x] == x) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return;
    if(rank[x] < rank[y]) {
      par[x] = y; s[y] = s[x] + s[y];
    } else {
      par[y] = x; s[x] = s[x] + s[y];
      if( rank[x] == rank[y] ) rank[x]++;
    }
  }
  int size(int x) { return s[find(x)];}
  bool same(int x, int y) { return find(x) == find(y);}
  int group(int n) {
    REP(i, n) used[find(i)] = true;
    int ret = 0;
    REP(i, n) if(used[i]) ret++;
    return ret;
  }
};
UnionFind uf;

VI a[10010];
signed main(void)
{
  while(true) {
    int n, m;
    cin >> n >> m;
    if(!n) break;
    REP(i, m) {
      int x, y, z;
      cin >> x >> y >> z;
      x--, y--;
      a[i] = {z, x, y};
    }
    sort(a, a+m);
    if(n == 2) {
      cout << a[0][0] << endl;
      continue;
    }

    uf.init(n);
    VI v;
    REP(i, m) {
      if(!uf.same(a[i][1], a[i][2])) {
        v.PB(a[i][0]);
        uf.unite(a[i][1], a[i][2]);
      }
    }
    sort(ALL(v));
    cout << v[n/2-1] << endl;
  }

  return 0;
}
```