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 PYTHON):

Solution:
import sys
input = sys.stdin.readline

"""
Given a directed graph, find_SCC returns a list of lists containing 
the strongly connected components in topological order.
Note that this implementation can be also be used to check if a directed graph is a
DAG, and in that case it can be used to find the topological ordering of the nodes.
"""

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]

t = int(input())
for cas in range(t):
    input()
    n,m = [int(x) for x in input().split()]
    coupl = [[] for _ in range(n)]
    self_loop = [0] * n
    for _ in range(m):
        u,v = [int(x) - 1 for x in input().split()]
        if u != v:
            coupl[u].append(v)
        else:
            self_loop[u] = 1

    SCC = find_SCC(coupl)
    k = len(SCC)

    inSCC = [-1] * n
    for j,scc in enumerate(SCC):
        for node in scc:
            inSCC[node] = j

    DP = [0] * k
    DP[inSCC[0]] = 1
    is_infinite = [+(len(scc) > 1 or self_loop[scc[0]]) for scc in SCC]

    inf = 10**9

    for scc in SCC:
        for node in scc:
            scc1 = inSCC[node]
            if DP[scc1]:
                for nei in coupl[node]:
                    scc2 = inSCC[nei]
                    DP[scc2] = min(inf, DP[scc2] + DP[scc1])
                    is_infinite[scc2] |= is_infinite[scc1]

    ans = []
    for node in range(n):
        dp = DP[inSCC[node]]
        infi = is_infinite[inSCC[node]]

        if dp == 0:
            ans.append(0)
        elif infi:
            ans.append(-1)
        elif dp == 1:
            ans.append(1)
        else:
            ans.append(2)

    print ' '.join(str(x) for x in ans)