#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
//const ll mod = 1000000007;
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<ll> Size;
    UnionFind(int n = 1) {
        init(n);
    }

    void init(int n = 1) {
        par.resize(n + 1); rank.resize(n + 1); Size.resize(n + 1);
        for (int i = 0; i <= n; ++i) par[i] = i, rank[i] = 0, Size[i] = 1;
    }

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

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        Size[x] += Size[y];
        return true;
    }

    ll size(int x){
        return Size[root(x)];
    }
};

int N, M;

int s[10000], t[10000], c[10000];

typedef pair<int, i_i> iii;

bool f(int border) {
    UnionFind uni(N);
    vector<iii> v;
    //cerr << "--------" << border << "---------" << endl;
    for(int i = 0; i < M; i++) {
        v.push_back({0, {s[i], t[i]}});
        if(c[i] > border) v.back().first = 1;
    }
    sort(v.begin(), v.end());
    int cost = 0;
    for(auto tmp : v) {
        //cerr << tmp.first << " " << tmp.second.first << " " << tmp.second.second << endl;
        if(uni.merge(tmp.second.first, tmp.second.second)) {
            cost += tmp.first;
        }
    }
    return cost < N / 2;
}

void solve() {
    for(int i = 0; i < M; i++) cin >> s[i] >> t[i] >> c[i];
    int ok = 1e9;
    int ng = 0;
    while(abs(ok - ng) > 1) {
        int mid = (ok + ng) / 2;
        if(f(mid)) ok = mid;
        else ng = mid;
    }
    cout << ok << endl;
}

int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    while(cin >> N >> M) {
        if(N == 0) break;
        solve();
    }
    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;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
//const ll mod = 1000000007;
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<ll> Size;
    UnionFind(int n = 1) {
        init(n);
    }

    void init(int n = 1) {
        par.resize(n + 1); rank.resize(n + 1); Size.resize(n + 1);
        for (int i = 0; i <= n; ++i) par[i] = i, rank[i] = 0, Size[i] = 1;
    }

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

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        Size[x] += Size[y];
        return true;
    }

    ll size(int x){
        return Size[root(x)];
    }
};

int N, M;

int s[10000], t[10000], c[10000];

typedef pair<int, i_i> iii;

bool f(int border) {
    UnionFind uni(N);
    vector<iii> v;
    //cerr << "--------" << border << "---------" << endl;
    for(int i = 0; i < M; i++) {
        v.push_back({0, {s[i], t[i]}});
        if(c[i] > border) v.back().first = 1;
    }
    sort(v.begin(), v.end());
    int cost = 0;
    for(auto tmp : v) {
        //cerr << tmp.first << " " << tmp.second.first << " " << tmp.second.second << endl;
        if(uni.merge(tmp.second.first, tmp.second.second)) {
            cost += tmp.first;
        }
    }
    return cost < N / 2;
}

void solve() {
    for(int i = 0; i < M; i++) cin >> s[i] >> t[i] >> c[i];
    int ok = 1e9;
    int ng = 0;
    while(abs(ok - ng) > 1) {
        int mid = (ok + ng) / 2;
        if(f(mid)) ok = mid;
        else ng = mid;
    }
    cout << ok << endl;
}

int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    while(cin >> N >> M) {
        if(N == 0) break;
        solve();
    }
    return 0;
}

```