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>
#pragma GCC optimize(2)
using namespace std;
const double eps = 1e-8;
const long long inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
std::mt19937 rnd(233);
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
long long tt;
namespace FastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
template <class T, class... U>
bool read(T& h, U&... t) {
  return read(h) && read(t...);
}
};  // namespace FastIO
inline long long rd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void out(long long a) {
  if (a < 0) putchar('-'), a = -a;
  if (a >= 10) out(a / 10);
  putchar(a % 10 + '0');
}
long long ksm(long long a, long long b, long long m) {
  long long ret = 1ll;
  while (b) {
    if (b & 1) ret *= a, ret %= m;
    a *= a, a %= m, b >>= 1;
  }
  return ret;
}
long long ksm(long long a, long long b) {
  long long ret = 1ll;
  while (b) {
    if (b & 1) ret *= a;
    a *= a, b >>= 1;
  }
  return ret;
}
long long inv(long long x, long long m) { return ksm(x, m - 2, m); }
const long long N = 2e5 + 400;
const long long INF = 1e17;
long long n, m, S, T;
struct Dinic {
  long long head[N], dep[N];
  long long ecnt;
  struct edge {
    long long v, next;
    long long flow;
  } e[N * 10];
  void init() {
    memset(head, -1, sizeof head);
    ecnt = 0;
  }
  void addedge(long long u, long long v, long long flow) {
    e[ecnt].v = v;
    e[ecnt].flow = flow;
    e[ecnt].next = head[u];
    head[u] = ecnt++;
    e[ecnt].v = u;
    e[ecnt].flow = 0;
    e[ecnt].next = head[v];
    head[v] = ecnt++;
  }
  bool BFS() {
    memset(dep, 0, sizeof dep);
    queue<long long> Q;
    Q.push(S);
    dep[S] = 1;
    while (!Q.empty()) {
      long long u = Q.front();
      Q.pop();
      for (long long i = head[u]; ~i; i = e[i].next) {
        if (e[i].flow && !dep[e[i].v]) {
          dep[e[i].v] = dep[u] + 1;
          Q.push(e[i].v);
        }
      }
    }
    return dep[T];
  }
  long long DFS(long long u, long long f) {
    if (u == T || f == 0) return f;
    long long w, used = 0;
    for (long long i = head[u]; ~i; i = e[i].next) {
      if (e[i].flow && dep[e[i].v] == dep[u] + 1) {
        w = DFS(e[i].v, min(f, e[i].flow));
        e[i].flow -= w;
        e[i ^ 1].flow += w;
        used += w;
        f -= w;
      }
    }
    if (!used) dep[u] = -1;
    return used;
  }
  long long maxflow() {
    long long ans = 0;
    while (BFS()) {
      ans += DFS(S, INF);
    }
    return ans;
  }
} dinic;
vector<pair<pair<long long, long long>, long long>> vec;
bool check(long long x) {
  dinic.init();
  S = 0, T = n * 2 + 50;
  for (long long i = 1; i <= n; i++)
    dinic.addedge(S, i, 1), dinic.addedge(i + n, T, 1);
  for (auto i : vec) {
    if (i.second <= x) {
      dinic.addedge(i.first.first, i.first.second, 1);
    }
  }
  return dinic.maxflow() >= n;
}
void run() {
  n = rd(), m = rd();
  while (m--) {
    long long u = rd(), v = rd() + n, w = rd();
    vec.push_back(make_pair(make_pair(u, v), w));
  }
  long long l = 1, r = 1e9, ans = -1;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (check(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  out(ans), puts("");
}
signed main() {
  run();
  return 0;
}