Problem: The pandemic is upon us, and the world is in shortage of the most important resource: toilet paper. As one of the best prepared nations for this crisis, BubbleLand promised to help all other world nations with this valuable resource. To do that, the country will send airplanes to other countries carrying toilet paper.

In BubbleLand, there are N toilet paper factories, and N airports. Because of how much it takes to build a road, and of course legal issues, every factory must send paper to only one airport, and every airport can only take toilet paper from one factory.

Also, a road can't be built between all airport-factory pairs, again because of legal issues. Every possible road has number d given, number of days it takes to build that road.

Your job is to choose N factory-airport pairs, such that if the country starts building all roads at the same time, it takes the least amount of days to complete them.

Input

The first line contains two integers N (1 ≤ N ≤ 10^4) - number of airports/factories, and M (1 ≤ M ≤ 10^5) - number of available pairs to build a road between.

On next M lines, there are three integers u, v (1 ≤ u,v ≤ N), d (1 ≤ d ≤ 10^9) - meaning that you can build a road between airport u and factory v for d days.

Output

If there are no solutions, output -1. If there exists a solution, output the minimal number of days to complete all roads, equal to maximal d among all chosen roads.

Example

Input


3 5
1 2 1
2 3 2
3 3 3
2 1 4
2 2 5


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template <class t>
using vvc = vc<vc<t>>;
using pi = pair<int, int>;
using vi = vc<int>;
template <class t, class u>
ostream& operator<<(ostream& os, const pair<t, u>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class t>
ostream& operator<<(ostream& os, const vc<t>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
using uint = unsigned;
using ull = unsigned long long;
template <class t, size_t n>
ostream& operator<<(ostream& os, const array<t, n>& first) {
  return os << vc<t>(first.begin(), first.end());
}
template <int i, class T>
void print_tuple(ostream&, const T&) {}
template <int i, class T, class H, class... Args>
void print_tuple(ostream& os, const T& t) {
  if (i) os << ",";
  os << get<i>(t);
  print_tuple<i + 1, T, Args...>(os, t);
}
template <class... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
  os << "{";
  print_tuple<0, tuple<Args...>, Args...>(os, t);
  return os << "}";
}
template <class t>
void print(t x, int suc = 1) {
  cout << x;
  if (suc == 1) cout << "\n";
  if (suc == 2) cout << " ";
}
ll read() {
  ll i;
  cin >> i;
  return i;
}
vi readvi(int n, int off = 0) {
  vi v(n);
  for (int i = int(0); i < int(n); i++) v[i] = read() + off;
  return v;
}
pi readpi(int off = 0) {
  int first, second;
  cin >> first >> second;
  return pi(first + off, second + off);
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = int(0); i < int(v.size()); i++)
    print(v[i], i == int(v.size()) - 1 ? suc : 2);
}
string readString() {
  string s;
  cin >> s;
  return s;
}
template <class T>
T sq(const T& t) {
  return t * t;
}
void yes(bool ex = true) {
  cout << "Yes"
       << "\n";
  if (ex) exit(0);
}
void no(bool ex = true) {
  cout << "No"
       << "\n";
  if (ex) exit(0);
}
void possible(bool ex = true) {
  cout << "Possible"
       << "\n";
  if (ex) exit(0);
}
void impossible(bool ex = true) {
  cout << "Impossible"
       << "\n";
  if (ex) exit(0);
}
constexpr ll ten(int n) { return n == 0 ? 1 : ten(n - 1) * 10; }
const ll infLL = LLONG_MAX / 3;
const int inf = INT_MAX / 2 - 100;
int topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
int topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
int botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
int botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
int popcount(signed t) { return __builtin_popcount(t); }
int popcount(ll t) { return __builtin_popcountll(t); }
bool ispow2(int i) { return i && (i & -i) == i; }
ll mask(int i) { return (ll(1) << i) - 1; }
bool inc(int first, int second, int c) {
  return first <= second && second <= c;
}
template <class t>
void mkuni(vc<t>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
ll rand_int(ll l, ll r) {
  static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<ll>(l, r)(gen);
}
template <class t>
void myshuffle(vc<t>& first) {
  for (int i = int(0); i < int(int(first.size())); i++)
    swap(first[i], first[rand_int(0, i)]);
}
template <class t>
int lwb(const vc<t>& v, const t& first) {
  return lower_bound(v.begin(), v.end(), first) - v.begin();
}
vvc<int> readGraph(int n, int m) {
  vvc<int> g(n);
  for (int i = int(0); i < int(m); i++) {
    int first, second;
    cin >> first >> second;
    first--;
    second--;
    g[first].push_back(second);
    g[second].push_back(first);
  }
  return g;
}
vvc<int> readTree(int n) { return readGraph(n, n - 1); }
template <class E>
struct bipartitematching {
  int n, m;
  vvc<E> g;
  vi dist, vis;
  vi to, ot;
  int sz;
  bipartitematching(int nn, int mm)
      : n(nn), m(mm), g(n), dist(n), vis(n), to(n, -1), ot(m, -1), sz(0) {}
  void ae(int first, E e) { g[first].push_back(e); }
  void bfs() {
    fill(dist.begin(), dist.end(), -1);
    vi q;
    int h = 0;
    auto rc = [&](int v, int d) {
      if (v == -1 || dist[v] != -1) return;
      dist[v] = d;
      q.push_back(v);
    };
    for (int i = int(0); i < int(n); i++)
      if (to[i] == -1) rc(i, 0);
    while (h < (int)q.size()) {
      int v = q[h++];
      for (auto e : g[v]) rc(ot[e], dist[v] + 1);
    }
  }
  bool dfs(int v) {
    if (vis[v]) return 0;
    vis[v] = 1;
    for (auto e : g[v]) {
      if (ot[e] == -1 || (dist[v] + 1 == dist[ot[e]] && dfs(ot[e]))) {
        to[v] = e;
        ot[e] = v;
        return 1;
      }
    }
    return 0;
  }
  void calc() {
    while (1) {
      bfs();
      fill(vis.begin(), vis.end(), 0);
      int f = 0;
      for (int i = int(0); i < int(n); i++)
        if (to[i] == -1)
          if (dfs(i)) f++;
      if (f == 0) break;
      sz += f;
    }
  }
};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  using T = tuple<int, int, int>;
  int n, m;
  cin >> n >> m;
  vc<T> es;
  for (int i = int(0); i < int(m); i++) {
    int first, second, c;
    cin >> first >> second >> c;
    first--;
    second--;
    es.emplace_back(c, first, second);
  }
  sort(es.begin(), es.end());
  int lw = 0, up = m + 1;
  while (up - lw > 1) {
    int mid = (lw + up) / 2;
    bipartitematching<int> bm(n, n);
    for (int i = int(0); i < int(mid); i++) {
      int c, first, second;
      tie(c, first, second) = es[i];
      bm.ae(first, second);
    }
    bm.calc();
    if (bm.sz == n)
      up = mid;
    else
      lw = mid;
  }
  int ans = -1;
  if (lw < m) ans = get<0>(es[lw]);
  print(ans);
}