#include<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=1000000007;

typedef tuple<ll,ll,ll> e;

vector<ll> uft;

ll parent(ll a){
    if(uft[a]==a){return a;}
    return uft[a]=parent(uft[a]);
}

int main(){
    ll n,m;
    while(cin>>n>>m){
        if(n==0){break;}
    uft.resize(n);
    for(int i=0;i<n;i++){uft[i]=i;}
    priority_queue<e,vector<e>,greater<e>> Q;
    for(int i=0;i<m;i++){
        ll a,b,c;
        cin>>a>>b>>c;
        a--; b--;
        Q.push(make_tuple(c,a,b));
    }
    ll ans=-1;
    ll count=0;
    while(!Q.empty()){
        e W=Q.top(); Q.pop();
        if(parent(get<1>(W))!=parent(get<2>(W))){
            count++;
            if(count==n/2){ans=get<0>(W);}
            uft[parent(get<1>(W))]=parent(get<2>(W));
        }
    }
    cout<<ans<<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<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=1000000007;

typedef tuple<ll,ll,ll> e;

vector<ll> uft;

ll parent(ll a){
    if(uft[a]==a){return a;}
    return uft[a]=parent(uft[a]);
}

int main(){
    ll n,m;
    while(cin>>n>>m){
        if(n==0){break;}
    uft.resize(n);
    for(int i=0;i<n;i++){uft[i]=i;}
    priority_queue<e,vector<e>,greater<e>> Q;
    for(int i=0;i<m;i++){
        ll a,b,c;
        cin>>a>>b>>c;
        a--; b--;
        Q.push(make_tuple(c,a,b));
    }
    ll ans=-1;
    ll count=0;
    while(!Q.empty()){
        e W=Q.top(); Q.pop();
        if(parent(get<1>(W))!=parent(get<2>(W))){
            count++;
            if(count==n/2){ans=get<0>(W);}
            uft[parent(get<1>(W))]=parent(get<2>(W));
        }
    }
    cout<<ans<<endl;
    }
    
    
    return 0;
}


```