Problem: You are given a directed graph G which can contain loops (edges from a vertex to itself). Multi-edges are absent in G which means that for all ordered pairs (u, v) exists at most one edge from u to v. Vertices are numbered from 1 to n.

A path from u to v is a sequence of edges such that:

  * vertex u is the start of the first edge in the path; 
  * vertex v is the end of the last edge in the path; 
  * for all pairs of adjacent edges next edge starts at the vertex that the previous edge ends on. 



We will assume that the empty sequence of edges is a path from u to u.

For each vertex v output one of four values:

  * 0, if there are no paths from 1 to v; 
  * 1, if there is only one path from 1 to v; 
  * 2, if there is more than one path from 1 to v and the number of paths is finite; 
  * -1, if the number of paths from 1 to v is infinite. 



Let's look at the example shown in the figure.

<image>

Then:

  * the answer for vertex 1 is 1: there is only one path from 1 to 1 (path with length 0); 
  * the answer for vertex 2 is 0: there are no paths from 1 to 2; 
  * the answer for vertex 3 is 1: there is only one path from 1 to 3 (it is the edge (1, 3)); 
  * the answer for vertex 4 is 2: there are more than one paths from 1 to 4 and the number of paths are finite (two paths: [(1, 3), (3, 4)] and [(1, 4)]); 
  * the answer for vertex 5 is -1: the number of paths from 1 to 5 is infinite (the loop can be used in a path many times); 
  * the answer for vertex 6 is -1: the number of paths from 1 to 6 is infinite (the loop can be used in a path many times). 

Input

The first contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow. Before each test case, there is an empty line.

The first line of the test case contains two integers n and m (1 ≤ n ≤ 4 ⋅ 10^5, 0 ≤ m ≤ 4 ⋅ 10^5) — numbers of vertices and edges in graph respectively. The next m lines contain edges descriptions. Each line contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n) — the start and the end of the i-th edge. The vertices of the graph are numbered from 1 to n. The given graph can contain loops (it is possible that a_i = b_i), but cannot contain multi-edges (it is not possible that a_i = a_j and b_i = b_j for i ≠ j).

The sum of n over all test cases does not exceed 4 ⋅ 10^5. Similarly, the sum of m over all test cases does not exceed 4 ⋅ 10^5.

Output

Output t lines. The i-th line should contain an answer for the i-th test case: a sequence of n integers from -1 to 2.

Example

Input


5

6 7
1 4
1 3
3 4
4 5
2 1
5 5
5 6

1 0

3 3
1 2
2 3
3 1

5 0

4 4
1 2
2 3
1 4
4 3


Output


1 0 1 2 -1 -1 
1 
-1 -1 -1 
1 0 0 0 0 
1 1 2 1 

Solution (in PYTHON3):

Solution:
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        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 = os.read(self._fd, max(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 = os.read(self._fd, max(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:
            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")

# ------------------- fast io --------------------
from math import ceil


def prod(a, mod=10 ** 9 + 7):
    ans = 1
    for each in a:
        ans = (ans * each) % mod
    return ans


def gcd(x, y):
    while y:
        x, y = y, x % y
    return x


def lcm(a, b): return a * b // gcd(a, b)


def binary(x, length=16):
    y = bin(x)[2:]
    return y if len(y) >= length else "0" * (length - len(y)) + y


def find_SCC(graph):
    SCC, S, P = [], [], []
    depth = [0] * len(graph)
    
    stack = list(range(len(graph)))
    while stack:
        node = stack.pop()
        if node < 0:
            d = depth[~node] - 1
            if P[-1] > d:
                SCC.append(S[d:])
                del S[d:], P[-1]
                for node in SCC[-1]:
                    depth[node] = -1
        elif depth[node] > 0:
            while P[-1] > depth[node]:
                P.pop()
        elif depth[node] == 0:
            S.append(node)
            P.append(len(S))
            depth[node] = len(S)
            stack.append(~node)
            stack += graph[node]
    return SCC[::-1]

for _ in range(int(input()) if True else 1):
    empty = input()
    n, m = map(int, input().split())
    # a, b = map(int, input().split())
    # c, d = map(int, input().split())
    # a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # s = input()
    graph = [[] for __ in range(n + 1)]
    self = [False] * (n + 1)
    for i in range(m):
        x, y = map(int, input().split())
        if x == y: self[x] = True
        graph[x] += [y]
    visited = [False] * (n + 1)
    multiple = [False] * (n + 1)
    queue = [1]
    visited[1] = True
    while queue:
        x = queue.pop()
        for y in graph[x]:
            if not visited[y]:
                visited[y] = True
                queue += [y]
            else:
                if not multiple[y]:
                    multiple[y] = True
                    queue += [y]
    
    infinite = [False] * (n + 1)
    for comp in find_SCC(graph):
        if len(comp) > 1 or (len(comp) == 1 and self[comp[0]]):
            if not infinite[comp[0]] and visited[comp[0]]:
                queue = comp
                for x in queue: infinite[x] = True
                while queue:
                    x = queue.pop()
                    for y in graph[x]:
                        if not infinite[y]:
                            infinite[y] = True
                            queue += [y]
                
    ans = []
    for i in range(1, n + 1):
        if infinite[i] and multiple[i]:
            ans += [-1]
        elif multiple[i]:
            ans += [2]
        elif visited[i]:
            ans += [1]
        else:
            ans += [0]
    print(*ans)