#include "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
template<class T>
using Table = vector<vector<T>>;
const ld eps=1e-9;

//// < "D:\D_Download\Visual Studio 2015\Projects\programing_contest_c++\Debug\a.txt"

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

struct edge {
  int f;
  int t;
  int c;
};
int main() {
  while (1) {
    int N, M; cin >> N >> M;
    if (!N)break;
    vector<edge>es(N);
    for(int i = 0; i < M; ++i) {
      int f, t, c; cin >> f >> t >> c;
      f--; t--;
      es.push_back(edge{ f,t,c });
    }
    UnionFind uf(N);
    sort(es.begin(), es.end(), [](const edge&l, const edge&r) {return l.c < r.c; });
    vector<int>nums;
    for (int e = 0; e < es.size(); ++e) {
      int f = es[e].f;
      int t = es[e].t;
      if (uf.root(f) == uf.root(t))continue;
      else {
        uf.unionSet(f, t);
        nums.push_back(es[e].c);
      }
    }
    int ans = nums[N / 2-1];
    
    cout << ans << 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 "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
template<class T>
using Table = vector<vector<T>>;
const ld eps=1e-9;

//// < "D:\D_Download\Visual Studio 2015\Projects\programing_contest_c++\Debug\a.txt"

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

struct edge {
  int f;
  int t;
  int c;
};
int main() {
  while (1) {
    int N, M; cin >> N >> M;
    if (!N)break;
    vector<edge>es(N);
    for(int i = 0; i < M; ++i) {
      int f, t, c; cin >> f >> t >> c;
      f--; t--;
      es.push_back(edge{ f,t,c });
    }
    UnionFind uf(N);
    sort(es.begin(), es.end(), [](const edge&l, const edge&r) {return l.c < r.c; });
    vector<int>nums;
    for (int e = 0; e < es.size(); ++e) {
      int f = es[e].f;
      int t = es[e].t;
      if (uf.root(f) == uf.root(t))continue;
      else {
        uf.unionSet(f, t);
        nums.push_back(es[e].c);
      }
    }
    int ans = nums[N / 2-1];
    
    cout << ans << endl;
  }
  return 0;
}
```