#include <iostream>
#include <vector>
#include <algorithm>
#include <array>
#include <set>
#include <map>
#include <queue>
#include <tuple>
#include <unordered_set>
#include <unordered_map>
#include <functional>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define whole(f,x,...) ([&](decltype((x)) y) { return (f)(begin(y), end(y), ## __VA_ARGS__); })(x)
using namespace std;
struct disjoint_sets {
    vector<int> xs;
    explicit disjoint_sets(size_t n) : xs(n, -1) {}
    bool is_root(int i) { return xs[i] < 0; }
    int find_root(int i) { return is_root(i) ? i : (xs[i] = find_root(xs[i])); }
    int set_size(int i) { return - xs[find_root(i)]; }
    int union_sets(int i, int j) {
        i = find_root(i); j = find_root(j);
        if (i != j) {
            if (set_size(i) < set_size(j)) swap(i,j);
            xs[i] += xs[j];
            xs[j] = i;
        }
        return i;
    }
    bool is_same(int i, int j) { return find_root(i) == find_root(j); }
};
struct edge_t { int s, t, c; };
bool operator < (edge_t const & a, edge_t const & b) { return a.c < b.c; } // strict weak ordering
int main() {
    while (true) {
        int n, m; cin >> n >> m;
        if (n == 0 and m == 0) break;
        vector<edge_t> es(m);
        repeat (i,m) {
            cin >> es[i].s >> es[i].t >> es[i].c;
            -- es[i].s;
            -- es[i].t;
        }
        whole(sort, es);
        disjoint_sets t(n);
        vector<int> acc;
        for (edge_t e : es) {
            if (not t.is_same(e.s, e.t)) {
                t.union_sets(e.s, e.t);
                acc.push_back(e.c);
            }
        }
        whole(sort, acc);
        double ans = (n-1) % 2 == 0 ? (acc[(n-1)/2] + acc[(n-1)/2+1]) / 2.0 : acc[(n-1)/2];
        cout << ans << endl;
    }
    return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <vector>
#include <algorithm>
#include <array>
#include <set>
#include <map>
#include <queue>
#include <tuple>
#include <unordered_set>
#include <unordered_map>
#include <functional>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define whole(f,x,...) ([&](decltype((x)) y) { return (f)(begin(y), end(y), ## __VA_ARGS__); })(x)
using namespace std;
struct disjoint_sets {
    vector<int> xs;
    explicit disjoint_sets(size_t n) : xs(n, -1) {}
    bool is_root(int i) { return xs[i] < 0; }
    int find_root(int i) { return is_root(i) ? i : (xs[i] = find_root(xs[i])); }
    int set_size(int i) { return - xs[find_root(i)]; }
    int union_sets(int i, int j) {
        i = find_root(i); j = find_root(j);
        if (i != j) {
            if (set_size(i) < set_size(j)) swap(i,j);
            xs[i] += xs[j];
            xs[j] = i;
        }
        return i;
    }
    bool is_same(int i, int j) { return find_root(i) == find_root(j); }
};
struct edge_t { int s, t, c; };
bool operator < (edge_t const & a, edge_t const & b) { return a.c < b.c; } // strict weak ordering
int main() {
    while (true) {
        int n, m; cin >> n >> m;
        if (n == 0 and m == 0) break;
        vector<edge_t> es(m);
        repeat (i,m) {
            cin >> es[i].s >> es[i].t >> es[i].c;
            -- es[i].s;
            -- es[i].t;
        }
        whole(sort, es);
        disjoint_sets t(n);
        vector<int> acc;
        for (edge_t e : es) {
            if (not t.is_same(e.s, e.t)) {
                t.union_sets(e.s, e.t);
                acc.push_back(e.c);
            }
        }
        whole(sort, acc);
        double ans = (n-1) % 2 == 0 ? (acc[(n-1)/2] + acc[(n-1)/2+1]) / 2.0 : acc[(n-1)/2];
        cout << ans << endl;
    }
    return 0;
}
```