#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

template <class T>
struct Kruskal {
  struct UnionFind {
    int N;
    vector<int> sz_tree, parent;
    UnionFind() {}
    UnionFind(int N) : N(N), sz_tree(N, 1), parent(N) {
      iota(parent.begin(), parent.end(), 0);
    }
    int root(int x) {
      return parent[x] == x ? x : parent[x] = root(parent[x]);
    }
    void merge(int x, int y) {
      if ((x = root(x)) == (y = root(y))) return;
      if (sz_tree[x] < sz_tree[y]) swap(x, y);
      sz_tree[x] += sz_tree[y];
      parent[y] = x;
    }
    bool same(int x, int y) {
      return root(x) == root(y);
    }
    int size(int x) {
      return sz_tree[root(x)];
    }
  };
  struct edge {
    int from, to; T cost;
    edge() {}
    edge(int f, int t, T c) : from(f), to(t), cost(c) {}
  };
 
  int N;
  vector<edge> es;
  vector<edge> use;
  Kruskal() {}
  Kruskal(int N) : N(N) {}
 
  void add_edge(int from, int to, T cost) {
    es.emplace_back(from, to, cost);
  }
  T mst() {
    T res = 0;
    UnionFind uf(N);
    sort(es.begin(), es.end(), [&](edge a, edge b) {
      return a.cost < b.cost;
    });
    for (auto &e : es) {
      if (uf.same(e.from, e.to)) continue;
      uf.merge(e.from, e.to);
      use.emplace_back(e);
      res += e.cost;
    }
    return res;
  }
};

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  int n, m;
  while (cin >> n >> m, n) {
    Kruskal<int> K(n);
    for (int i = 0; i < m; i++) {
      int s, t, c;
      cin >> s >> t >> c;
      s--, t--;
      K.add_edge(s, t, c);
      K.add_edge(t, s, c);
    }
    K.mst();
    cout << K.use[(n - 1) / 2].cost << endl;
  }

  return 0;
}

### 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
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

template <class T>
struct Kruskal {
  struct UnionFind {
    int N;
    vector<int> sz_tree, parent;
    UnionFind() {}
    UnionFind(int N) : N(N), sz_tree(N, 1), parent(N) {
      iota(parent.begin(), parent.end(), 0);
    }
    int root(int x) {
      return parent[x] == x ? x : parent[x] = root(parent[x]);
    }
    void merge(int x, int y) {
      if ((x = root(x)) == (y = root(y))) return;
      if (sz_tree[x] < sz_tree[y]) swap(x, y);
      sz_tree[x] += sz_tree[y];
      parent[y] = x;
    }
    bool same(int x, int y) {
      return root(x) == root(y);
    }
    int size(int x) {
      return sz_tree[root(x)];
    }
  };
  struct edge {
    int from, to; T cost;
    edge() {}
    edge(int f, int t, T c) : from(f), to(t), cost(c) {}
  };
 
  int N;
  vector<edge> es;
  vector<edge> use;
  Kruskal() {}
  Kruskal(int N) : N(N) {}
 
  void add_edge(int from, int to, T cost) {
    es.emplace_back(from, to, cost);
  }
  T mst() {
    T res = 0;
    UnionFind uf(N);
    sort(es.begin(), es.end(), [&](edge a, edge b) {
      return a.cost < b.cost;
    });
    for (auto &e : es) {
      if (uf.same(e.from, e.to)) continue;
      uf.merge(e.from, e.to);
      use.emplace_back(e);
      res += e.cost;
    }
    return res;
  }
};

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  int n, m;
  while (cin >> n >> m, n) {
    Kruskal<int> K(n);
    for (int i = 0; i < m; i++) {
      int s, t, c;
      cin >> s >> t >> c;
      s--, t--;
      K.add_edge(s, t, c);
      K.add_edge(t, s, c);
    }
    K.mst();
    cout << K.use[(n - 1) / 2].cost << endl;
  }

  return 0;
}

```