#include <bits/stdc++.h>
using namespace std;

class UnionFind {
private:
  vector<int> par;
  vector<int> rank;

public:
  UnionFind(int n) {
    par.resize(n);
    rank.resize(n);
    for(int i = 0; i < n; ++i) {
      par[i] = i;
      rank[i] = 0;
    }
  }
  
  int find(int x) {
    if(par[x] == x) {
      return x;
    } else  {
      return par[x] = find(par[x]);
    }
  }

  bool same(int x, int y) {
    return find(x) == find(y);
  }
  
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return;
    
    if(rank[x] < rank[y]) {
      par[x] = y;
    } else {
      par[y] = x;
      if(rank[x] == rank[y]) ++rank[x];
    }
  }
};

struct Edge {
  int a, b, c;
  bool operator < (const Edge &e) const {
    return c < e.c;
  }
};

int main() {
  for(int n, m; cin >> n >> m && (n|m);) {
    vector<Edge> es;
    while(m--) {
      int a, b, c;
      cin >> a >> b >> c;
      --a; --b;
      es.push_back((Edge){a,b,c});
    }
    sort(es.begin(), es.end());
    UnionFind uf(n);
    int cnt = 0;
    for(int i = 0; i < es.size(); ++i) {
      if(uf.same(es[i].a, es[i].b)) continue;
      uf.unite(es[i].a, es[i].b);
      if(++cnt == n / 2) {
        cout << es[i].c << endl;
        break;
      }
    }
  }
  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
#include <bits/stdc++.h>
using namespace std;

class UnionFind {
private:
  vector<int> par;
  vector<int> rank;

public:
  UnionFind(int n) {
    par.resize(n);
    rank.resize(n);
    for(int i = 0; i < n; ++i) {
      par[i] = i;
      rank[i] = 0;
    }
  }
  
  int find(int x) {
    if(par[x] == x) {
      return x;
    } else  {
      return par[x] = find(par[x]);
    }
  }

  bool same(int x, int y) {
    return find(x) == find(y);
  }
  
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return;
    
    if(rank[x] < rank[y]) {
      par[x] = y;
    } else {
      par[y] = x;
      if(rank[x] == rank[y]) ++rank[x];
    }
  }
};

struct Edge {
  int a, b, c;
  bool operator < (const Edge &e) const {
    return c < e.c;
  }
};

int main() {
  for(int n, m; cin >> n >> m && (n|m);) {
    vector<Edge> es;
    while(m--) {
      int a, b, c;
      cin >> a >> b >> c;
      --a; --b;
      es.push_back((Edge){a,b,c});
    }
    sort(es.begin(), es.end());
    UnionFind uf(n);
    int cnt = 0;
    for(int i = 0; i < es.size(); ++i) {
      if(uf.same(es[i].a, es[i].b)) continue;
      uf.unite(es[i].a, es[i].b);
      if(++cnt == n / 2) {
        cout << es[i].c << endl;
        break;
      }
    }
  }
  return 0;
}
```