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 PYTHON3):

Solution:
import os
import sys
from io import BytesIO, IOBase
 
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        import os
        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
 
def getInts():
    return [int(s) for s in input().split()]
 
 
from collections import deque
 
def maximum_matching(graph):
    """Find a maximum unweighted matching in a bipartite graph. The input
    must be a dictionary mapping vertices in one partition to sets of
    vertices in the other partition. Return a dictionary mapping
    vertices in one partition to their matched vertex in the other.
 
    """
    # The two partitions of the graph.
    U = set(graph.keys())
    V = set.union(*graph.values())
 
    # A distinguished vertex not belonging to the graph.
    nil = object()
 
    # Current pairing for vertices in each partition.
    pair_u = dict.fromkeys(U, nil)
    pair_v = dict.fromkeys(V, nil)
    if len(pair_u) != len(pair_v):
        return []
 
    # Distance to each vertex along augmenting paths.
    dist = {}
    inf = (1 << 28)
 
    def bfs():
        # Breadth-first search of the graph starting with unmatched
        # vertices in U and following "augmenting paths" alternating
        # between edges from U->V not in the matching and edges from
        # V->U in the matching. This partitions the vertexes into
        # layers according to their distance along augmenting paths.
        queue = deque()
        for u in U:
            if pair_u[u] is nil:
                dist[u] = 0
                queue.append(u)
            else:
                dist[u] = inf
        dist[nil] = inf
        while queue:
            u = queue.popleft()
            if dist[u] < dist[nil]:
                # Follow edges from U->V not in the matching.
                for v in graph[u]:
                    # Follow edge from V->U in the matching, if any.
                    uu = pair_v[v]
                    if dist[uu] == inf:
                        dist[uu] = dist[u] + 1
                        queue.append(uu)
        return dist[nil] is not inf
 
    def bootstrap(f, stack=[]):
        def wrappedfunc(*args, **kwargs):
            if stack:
                return f(*args, **kwargs)
            else:
                to = f(*args, **kwargs)
                while True:
                    if type(to) is GeneratorType:
                        stack.append(to)
                        to = next(to)
                    else:
                        stack.pop()
                        if not stack:
                            break
                        to = stack[-1].send(to)
                return to
        return wrappedfunc
 
    @bootstrap
    def dfs(u):
        # Depth-first search along "augmenting paths" starting at
        # u. If we can find such a path that goes all the way from
        # u to nil, then we can swap matched/unmatched edges all
        # along the path, getting one additional matched edge
        # (since the path must have an odd number of edges).
        if u is not nil:
            for v in graph[u]:
                uu = pair_v[v]
                if dist[uu] == dist[u] + 1: # uu in next layer
                    if (dfs(uu)):
                        # Found augmenting path all the way to nil. In
                        # this path v was matched with uu, so after
                        # swapping v must now be matched with u.
                        pair_v[v] = u
                        pair_u[u] = v
                        yield True
            dist[u] = inf
            yield False
        yield True
 
    while bfs():
        for u in U:
            if pair_u[u] is nil:
                dfs(u)
    return {u: v for u, v in pair_u.items() if v is not nil}
 
N, M = getInts()
max_edge = 0
edges = []
 
u_set = set()
v_set = set()
 
while M:
    u, v, d = getInts()
    u_set.add(u)
    v_set.add(v)
    edges.append((u,v,d))
    max_edge = max(d,max_edge)
    M -= 1
 
l = 0
r = max_edge+1
ans = 2*(10**9)
import collections
if len(u_set) != N or len(v_set) != N:
    print(-1)

else:
 
    while (l <= r):
        m = (l + r) >> 1
        G = collections.defaultdict(set)
        for u,v,d in edges:
            if d <= m:
                G[u].add(v+N)
        #print(m,G)
        if not G:
            l = m + 1
            continue
        matching = maximum_matching(G)
        if (len(matching) == N):
            ans = m
            r = m - 1
        else:
            l = m + 1
        
    if (ans == 2*(10**9)):
        print(-1)
    else:
        print(ans)