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

typedef tuple<int, int, int> T;

const int MAX_N = 1010;

int n, m;

int par[MAX_N];
int ranks[MAX_N];
int ufSize;

void init(int n){
    ufSize = n;
    for(int i=0; i<n; i++){
        par[i] = i;
        ranks[i] = 0;
    }
}

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

bool same(int x, int y){
    return find(x) == find(y);
}

void unite(int x, int y){
    x = find(x);
    y = find(y);
    if(x == y) return;
    ufSize--;

    if(ranks[x] < ranks[y]){
        par[x] = y;
    }
    else{
        par[y] = x;
    }
    
    if(ranks[x] == ranks[y]){
        ranks[x]++;
    }
}

int kruskal(vector<T> &edge){
    sort(edge.begin(), edge.end());
    init(n);
    int cnt = 0;
    for(int i=0; i<edge.size(); i++){
        int cost, cv, nv;
        tie(cost, cv, nv) = edge[i];

        if(!same(cv, nv)){
            cnt++;

            if(cnt == n / 2){
                return cost;
            }

            unite(cv, nv);
        }
    }
}

int main(){
    while(1){
        cin>>n>>m;
        if(n == 0) return 0;

        vector<T> edge;
        for(int i=0; i<m; i++){
            int s, t, c; cin>>s>>t>>c;
            s--, t--;
            edge.emplace_back(c, s, t);
        }

        cout << kruskal(edge) << endl;
    }
}

### Prompt

Please formulate 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>
using namespace std;

typedef tuple<int, int, int> T;

const int MAX_N = 1010;

int n, m;

int par[MAX_N];
int ranks[MAX_N];
int ufSize;

void init(int n){
    ufSize = n;
    for(int i=0; i<n; i++){
        par[i] = i;
        ranks[i] = 0;
    }
}

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

bool same(int x, int y){
    return find(x) == find(y);
}

void unite(int x, int y){
    x = find(x);
    y = find(y);
    if(x == y) return;
    ufSize--;

    if(ranks[x] < ranks[y]){
        par[x] = y;
    }
    else{
        par[y] = x;
    }
    
    if(ranks[x] == ranks[y]){
        ranks[x]++;
    }
}

int kruskal(vector<T> &edge){
    sort(edge.begin(), edge.end());
    init(n);
    int cnt = 0;
    for(int i=0; i<edge.size(); i++){
        int cost, cv, nv;
        tie(cost, cv, nv) = edge[i];

        if(!same(cv, nv)){
            cnt++;

            if(cnt == n / 2){
                return cost;
            }

            unite(cv, nv);
        }
    }
}

int main(){
    while(1){
        cin>>n>>m;
        if(n == 0) return 0;

        vector<T> edge;
        for(int i=0; i<m; i++){
            int s, t, c; cin>>s>>t>>c;
            s--, t--;
            edge.emplace_back(c, s, t);
        }

        cout << kruskal(edge) << endl;
    }
}

```