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

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define repr(i,n) for(int i=(int)(n-1);i>=0;i--)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout << #x" = " << x << endl
#define print(x) cout << x << endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<int, P> PIP;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;
typedef set<int> S;

#define INF INT_MAX/3
#define MAX_N 1000000001

int par[1000];

struct Edge {
  int s, t, c;
};

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

int find(int v) {
  if (v == par[v]) return v;
  return par[v] = find(par[v]);
}

void unite(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;

  par[u] = v;
}

int main(){
  cin.sync_with_stdio(false);
  int n, m;
  while(cin >> n >> m, n | m) {
    iota(par, par + n, 0);
    vector<Edge> e(m);
    rep(i, m) {
      cin >> e[i].s >> e[i].t >> e[i].c;
      e[i].s--, e[i].t--;
    }

    sort(all(e));

    vector<int> cost;
    rep(i, m) {
      if (find(e[i].s) != find(e[i].t)) {
        unite(e[i].s, e[i].t);
        cost.pb(e[i].c);
      }
    }

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

  return 0;
}
### Prompt

Please create a solution in cpp 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;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define repr(i,n) for(int i=(int)(n-1);i>=0;i--)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout << #x" = " << x << endl
#define print(x) cout << x << endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<int, P> PIP;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;
typedef set<int> S;

#define INF INT_MAX/3
#define MAX_N 1000000001

int par[1000];

struct Edge {
  int s, t, c;
};

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

int find(int v) {
  if (v == par[v]) return v;
  return par[v] = find(par[v]);
}

void unite(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;

  par[u] = v;
}

int main(){
  cin.sync_with_stdio(false);
  int n, m;
  while(cin >> n >> m, n | m) {
    iota(par, par + n, 0);
    vector<Edge> e(m);
    rep(i, m) {
      cin >> e[i].s >> e[i].t >> e[i].c;
      e[i].s--, e[i].t--;
    }

    sort(all(e));

    vector<int> cost;
    rep(i, m) {
      if (find(e[i].s) != find(e[i].t)) {
        unite(e[i].s, e[i].t);
        cost.pb(e[i].c);
      }
    }

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

  return 0;
}
```