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

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pi;
typedef pair<int,pi> edge;

struct UF{
    int n;
    //??£?????£????????????????????????,????????£???????????§??£?????????????????°
    vector<int> d;
    UF() {}
    UF(int N):n(N), d(N,-1){}
    int root(int v){
        if(d[v]<0) return v;
        return d[v]=root(d[v]);
    }
    bool unite(int X,int Y){
        X=root(X); Y=root(Y);
        if(X==Y) return false;
        if(size(X) < size(Y)) swap(X,Y);
        d[X]+=d[Y];
        d[Y]=X;
        return true;
    }
    int size(int v){ return -d[root(v)]; }
    bool same(int X,int Y){ return root(X)==root(Y); }
};

int main()
{
    int n,m;
    while(scanf(" %d %d", &n, &m),n)
    {
        vector<edge> e(m);
        rep(i,m)
        {
            int s,t,c;
            scanf(" %d %d %d", &s, &t, &c);
            --s;
            --t;
            e[i] = edge(c,pi(s,t));
        }

        sort(all(e));

        UF uf(n);
        vector<int> used;
        rep(i,m)
        {
            int a=e[i].se.fi, b=e[i].se.se;
            if(!uf.same(a,b))
            {
                uf.unite(a,b);
                used.pb(e[i].fi);
            }
        }

        printf("%d\n", used[n/2-1]);
    }
    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;

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pi;
typedef pair<int,pi> edge;

struct UF{
    int n;
    //??£?????£????????????????????????,????????£???????????§??£?????????????????°
    vector<int> d;
    UF() {}
    UF(int N):n(N), d(N,-1){}
    int root(int v){
        if(d[v]<0) return v;
        return d[v]=root(d[v]);
    }
    bool unite(int X,int Y){
        X=root(X); Y=root(Y);
        if(X==Y) return false;
        if(size(X) < size(Y)) swap(X,Y);
        d[X]+=d[Y];
        d[Y]=X;
        return true;
    }
    int size(int v){ return -d[root(v)]; }
    bool same(int X,int Y){ return root(X)==root(Y); }
};

int main()
{
    int n,m;
    while(scanf(" %d %d", &n, &m),n)
    {
        vector<edge> e(m);
        rep(i,m)
        {
            int s,t,c;
            scanf(" %d %d %d", &s, &t, &c);
            --s;
            --t;
            e[i] = edge(c,pi(s,t));
        }

        sort(all(e));

        UF uf(n);
        vector<int> used;
        rep(i,m)
        {
            int a=e[i].se.fi, b=e[i].se.se;
            if(!uf.same(a,b))
            {
                uf.unite(a,b);
                used.pb(e[i].fi);
            }
        }

        printf("%d\n", used[n/2-1]);
    }
    return 0;
}
```