#include <bits/stdc++.h>

using namespace std;

struct UnionFind
{
  vector<int> data;
  UnionFind(int n):data(n, -1){}
  int size(int x){ return -data[find(x)]; }
  int find(int x){ return data[x] < 0 ? x : data[x] = find(data[x]); }
  bool same(int x, int y) { return find(x) == find(y); }
  int unite(int x, int y)
  {
    x = find(x), y = find(y);
    if(x != y) {
      if(data[x] < data[y]) swap(x, y);
      data[x] += data[y]; data[y] = x;
    }
    return -data[x];
  }
};

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

vector<edge> E;
int n, m;

int Kruskal()
{
  sort(E.begin(), E.end());
  UnionFind uf(n);
  int cnt = 0;
  for(int i = 0; i < E.size(); i++) {
    edge& e = E[i];
    if(!uf.same(e.u, e.v)) {
      uf.unite(e.u, e.v);
      if(++cnt == n/2) return e.cost;
    }
  }
}

int main()
{
  while(cin >> n >> m, n || m) {
    E.clear();
    for(int i = 0; i < m; i++) {
      int s, t, c;
      cin >> s >> t >> c;
      --s, --t;
      E.push_back(edge(s, t, c));
    }
    cout << Kruskal() << endl;
  }
  return 0;
}
### Prompt

Please formulate 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 <bits/stdc++.h>

using namespace std;

struct UnionFind
{
  vector<int> data;
  UnionFind(int n):data(n, -1){}
  int size(int x){ return -data[find(x)]; }
  int find(int x){ return data[x] < 0 ? x : data[x] = find(data[x]); }
  bool same(int x, int y) { return find(x) == find(y); }
  int unite(int x, int y)
  {
    x = find(x), y = find(y);
    if(x != y) {
      if(data[x] < data[y]) swap(x, y);
      data[x] += data[y]; data[y] = x;
    }
    return -data[x];
  }
};

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

vector<edge> E;
int n, m;

int Kruskal()
{
  sort(E.begin(), E.end());
  UnionFind uf(n);
  int cnt = 0;
  for(int i = 0; i < E.size(); i++) {
    edge& e = E[i];
    if(!uf.same(e.u, e.v)) {
      uf.unite(e.u, e.v);
      if(++cnt == n/2) return e.cost;
    }
  }
}

int main()
{
  while(cin >> n >> m, n || m) {
    E.clear();
    for(int i = 0; i < m; i++) {
      int s, t, c;
      cin >> s >> t >> c;
      --s, --t;
      E.push_back(edge(s, t, c));
    }
    cout << Kruskal() << endl;
  }
  return 0;
}
```