#include <iostream>
#include <vector>
#include <algorithm>
#define REP(i, a, n) for(int i = (a); i < n; i++)
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;

int N, M, S, T, C;
pair<int, pii> E[10000];
int a[1000];

int find(int i) { return a[i] == i ? i : (a[i] = find(a[i])); }
void unite(int i, int j) { a[find(i)] = find(j); }

int main(void) {
  while(cin >> N >> M, N || M) {
    REP(i, 0, M) {
      cin >> S >> T >> C;
      E[i] = pair<int, pii>(C, pii(S, T));
    }

    sort(E, E + M);

    vector<int> cost;
    REP(i, 0, N) a[i] = i;
    REP(i, 0, M) {
      S = E[i].se.fi;
      T = E[i].se.se;
      C = E[i].fi;
      if(find(S) != find(T)) {
        unite(S, T);
        cost.push_back(C);
      }
    }

    cout << cost[cost.size() / 2] << endl;
  }

  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 <iostream>
#include <vector>
#include <algorithm>
#define REP(i, a, n) for(int i = (a); i < n; i++)
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;

int N, M, S, T, C;
pair<int, pii> E[10000];
int a[1000];

int find(int i) { return a[i] == i ? i : (a[i] = find(a[i])); }
void unite(int i, int j) { a[find(i)] = find(j); }

int main(void) {
  while(cin >> N >> M, N || M) {
    REP(i, 0, M) {
      cin >> S >> T >> C;
      E[i] = pair<int, pii>(C, pii(S, T));
    }

    sort(E, E + M);

    vector<int> cost;
    REP(i, 0, N) a[i] = i;
    REP(i, 0, M) {
      S = E[i].se.fi;
      T = E[i].se.se;
      C = E[i].fi;
      if(find(S) != find(T)) {
        unite(S, T);
        cost.push_back(C);
      }
    }

    cout << cost[cost.size() / 2] << endl;
  }

  return 0;
}
```