#include <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

struct UnionFind {
  vector<int> parent;
  UnionFind (int n) { parent.assign(n, -1); }
  int root(int x) { return parent[x] < 0 ? x : parent[x] = root(parent[x]); }
  bool merge(int x, int y) {
    x = root(x); y = root(y);
    if (x == y) return false;
    if (parent[y] < parent[x]) swap(x, y);
    if (parent[x] == parent[y]) --parent[x];
    parent[y] = x;
    return true;
  }
};

typedef int Weight;
struct Edge{
  int src, dest; Weight weight;
  bool operator < (const Edge &rhs) const {return weight > rhs.weight;}
};

typedef vector<Edge> Edges;

// Kruskal
// N, operator <
Edges kruskal(int V, Edges &es) {
  sort(es.rbegin(), es.rend());
  UnionFind uf(V);
  Edges res;
  REP(i, es.size()) {
    Edge e = es[i];
    if(uf.root(e.src) != uf.root(e.dest)) {
      uf.merge(e.src, e.dest);
      res.push_back(e);
    }
  }
  return res;
}

int main() {
  while(1){
    int n,m;
    cin>>n>>m;
    if(!n) break;
    Edges g;
    REP(i,m){
      int s,t,c;
      cin>>s>>t>>c;
      --s;--t;
      g.push_back((Edge){s,t,c});
      g.push_back((Edge){t,s,c});
    }
    auto res = kruskal(n, g);
    cout << res[n/2-1].weight << endl;
  }
  return 0;
}
### Prompt

Generate 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>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

struct UnionFind {
  vector<int> parent;
  UnionFind (int n) { parent.assign(n, -1); }
  int root(int x) { return parent[x] < 0 ? x : parent[x] = root(parent[x]); }
  bool merge(int x, int y) {
    x = root(x); y = root(y);
    if (x == y) return false;
    if (parent[y] < parent[x]) swap(x, y);
    if (parent[x] == parent[y]) --parent[x];
    parent[y] = x;
    return true;
  }
};

typedef int Weight;
struct Edge{
  int src, dest; Weight weight;
  bool operator < (const Edge &rhs) const {return weight > rhs.weight;}
};

typedef vector<Edge> Edges;

// Kruskal
// N, operator <
Edges kruskal(int V, Edges &es) {
  sort(es.rbegin(), es.rend());
  UnionFind uf(V);
  Edges res;
  REP(i, es.size()) {
    Edge e = es[i];
    if(uf.root(e.src) != uf.root(e.dest)) {
      uf.merge(e.src, e.dest);
      res.push_back(e);
    }
  }
  return res;
}

int main() {
  while(1){
    int n,m;
    cin>>n>>m;
    if(!n) break;
    Edges g;
    REP(i,m){
      int s,t,c;
      cin>>s>>t>>c;
      --s;--t;
      g.push_back((Edge){s,t,c});
      g.push_back((Edge){t,s,c});
    }
    auto res = kruskal(n, g);
    cout << res[n/2-1].weight << endl;
  }
  return 0;
}
```