#include <bits/stdc++.h>

using namespace std;

int par[1001];

void init(int n)
{
    memset(par, -1, sizeof(par));
}

int find(int x)
{
    if (par[x] < 0) return (x);
    return (par[x] = find(par[x]));
}

void merge(int x, int y)
{
    x = find(x);
    y = find(y);
    if (x == y) return;
    
    if (-par[x] < -par[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
}

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


int main()
{
    int n, m;
    
    while (scanf("%d %d", &n, &m) && n){
        vector<pair<int, pair<int, int> > > es(m);
        
        for (int i = 0; i < m; i++){
            int a, b, c;
            scanf("%d %d %d", &a, &b, &c);
            es[i] = make_pair(c, make_pair(a, b));
        }
        
        sort(es.begin(), es.end());
        
        int l = 1, r = 1000;
        
        while (l != r){
            int mid = l + r >> 1;
            init(n + 1);
            int ec = 0;
            int i = 0;
            while (ec != n - 1 && i != m){
                if (ec + 1 <= n / 2 && es[i].first > mid){
                    i++;
                    continue;
                }
                if (!same(es[i].second.first, es[i].second.second)){
                    merge(es[i].second.first, es[i].second.second);
                    ec++;
                }
                i++;
            }
            
            if (ec == n - 1) r = mid;
            else l = mid + 1;
        }
        
        printf("%d\n", l);
    }
    
    return (0);
}
### Prompt

In cpp, your task is to solve 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;

int par[1001];

void init(int n)
{
    memset(par, -1, sizeof(par));
}

int find(int x)
{
    if (par[x] < 0) return (x);
    return (par[x] = find(par[x]));
}

void merge(int x, int y)
{
    x = find(x);
    y = find(y);
    if (x == y) return;
    
    if (-par[x] < -par[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
}

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


int main()
{
    int n, m;
    
    while (scanf("%d %d", &n, &m) && n){
        vector<pair<int, pair<int, int> > > es(m);
        
        for (int i = 0; i < m; i++){
            int a, b, c;
            scanf("%d %d %d", &a, &b, &c);
            es[i] = make_pair(c, make_pair(a, b));
        }
        
        sort(es.begin(), es.end());
        
        int l = 1, r = 1000;
        
        while (l != r){
            int mid = l + r >> 1;
            init(n + 1);
            int ec = 0;
            int i = 0;
            while (ec != n - 1 && i != m){
                if (ec + 1 <= n / 2 && es[i].first > mid){
                    i++;
                    continue;
                }
                if (!same(es[i].second.first, es[i].second.second)){
                    merge(es[i].second.first, es[i].second.second);
                    ec++;
                }
                i++;
            }
            
            if (ec == n - 1) r = mid;
            else l = mid + 1;
        }
        
        printf("%d\n", l);
    }
    
    return (0);
}
```