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;
long long mod = 1000000007;
long long fact[200000];
long long dmod(long long x) { return ((x + mod) % mod); }
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
long long nCr(long long n, long long r) {
  if (r > n)
    return 0;
  else if (r < 0)
    return 0;
  else if (r == 0 || r == n)
    return 1;
  long long ans = fact[n];
  ans = (ans * power(fact[n - r], mod - 2)) % mod;
  ans = (ans * power(fact[r], mod - 2)) % mod;
  return ans;
}
struct FlowEdge {
  int v, u;
  long long cap, flow = 0;
  FlowEdge(int v, int u, long long cap) : v(v), u(u), cap(cap) {}
};
struct Dinic {
  const long long flow_inf = 1e18;
  vector<FlowEdge> edges;
  vector<vector<int>> adj;
  int n, m = 0;
  int s, t;
  vector<int> level, ptr;
  queue<int> q;
  Dinic(int n, int s, int t) : n(n), s(s), t(t) {
    adj.resize(n + 1);
    level.resize(n + 1);
    ptr.resize(n + 1);
  }
  void add_edge(int v, int u, long long cap) {
    edges.emplace_back(v, u, cap);
    edges.emplace_back(u, v, 0);
    adj[v].push_back(m);
    adj[u].push_back(m + 1);
    m += 2;
  }
  bool bfs() {
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (int id : adj[v]) {
        if (edges[id].cap - edges[id].flow < 1) continue;
        if (level[edges[id].u] != -1) continue;
        level[edges[id].u] = level[v] + 1;
        q.push(edges[id].u);
      }
    }
    return level[t] != -1;
  }
  long long dfs(int v, long long pushed) {
    if (pushed == 0) return 0;
    if (v == t) return pushed;
    for (int& cid = ptr[v]; cid < (int)adj[v].size(); cid++) {
      int id = adj[v][cid];
      int u = edges[id].u;
      if (level[v] + 1 != level[u] || edges[id].cap - edges[id].flow < 1)
        continue;
      long long tr = dfs(u, min(pushed, edges[id].cap - edges[id].flow));
      if (tr == 0) continue;
      edges[id].flow += tr;
      edges[id ^ 1].flow -= tr;
      return tr;
    }
    return 0;
  }
  long long flow() {
    long long f = 0;
    while (true) {
      fill(level.begin(), level.end(), -1);
      level[s] = 0;
      q.push(s);
      if (!bfs()) break;
      fill(ptr.begin(), ptr.end(), 0);
      while (long long pushed = dfs(s, flow_inf)) {
        f += pushed;
      }
    }
    return f;
  }
};
vector<long long> BIT;
long long query(long long i) {
  long long res = 0;
  while (i >= 0) {
    res += BIT[i];
    i = (i & (i + 1)) - 1;
  }
  return res;
}
void update(long long i, long long val) {
  while (i < BIT.size()) {
    BIT[i] += val;
    i = (i | (i + 1));
  }
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <class T, class H>
using umap = unordered_map<T, H, custom_hash>;
template <class T>
using uset = unordered_set<T, custom_hash>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<pair<pair<long long, long long>, long long>> vec;
  for (long long i = 0; i < m; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    vec.push_back({{a, b}, c});
  }
  long long low = 0, high = 1e9, ans = 1e9;
  while (low <= high) {
    long long mid = (high + low) / 2;
    Dinic d(2 * n + 2, 0, 2 * n + 1);
    for (long long i = 1; i <= n; i++) {
      d.add_edge(0, i, 1);
      d.add_edge(n + i, 2 * n + 1, 1);
    }
    for (int i = 0; i < vec.size(); i++) {
      if (vec[i].second > mid) continue;
      d.add_edge(vec[i].first.first, vec[i].first.second + n, 1);
    }
    long long res = d.flow();
    if (res == n) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  if (ans == 1e9)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}