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:
import sys

sys.setrecursionlimit(600000)


def dfs_1(v):
    q = [v, 0]
    while len(q) != 0:
        i = q.pop()
        v = q.pop()
        if i == 0:
            color[v] = 1
        if i == len(g[v]):
            color[v] = 2
            continue
        q.append(v)
        q.append(i + 1)
        w = g[v][i]
        if color[w] == 0:
            q.append(w)
            q.append(0)
        elif color[w] == 1:
            in_cycles.append(w)
        elif color[w] == 2:
            more_than_one.append(w)


def dfs_2(v):
    q = [v, 0]
    while len(q) != 0:
        i = q.pop()
        v = q.pop()
        if i == 0:
            color_tmp[v] = 1
        if i == len(g[v]):
            color_tmp[v] = 2
            continue
        q.append(v)
        q.append(i + 1)
        w = g[v][i]
        if color_tmp[w] == 0:
            q.append(w)
            q.append(0)


def fill_not_accessible(color, answer):
    for i, c in enumerate(color):
        if c == 0:
            answer[i] = 0


def fill_more_than_one(color, answer):
    for i, c in enumerate(color):
        if c == 2:
            answer[i] = 2


def fill_inf(color, answer):
    for i, c in enumerate(color):
        if c == 2:
            answer[i] = -1


t = int(input())

for _ in range(t):
    input()
    n, m = map(int, input().split())
    color = [0 for _ in range(n + 1)]
    g = {i: [] for i in range(n + 1)}
    in_cycles = []
    more_than_one = []
    answer = [1 for _ in range(n + 1)]
    for _ in range(m):
        a, b = map(int, input().split())
        g[a].append(b)

    dfs_1(1)

    fill_not_accessible(color, answer)

    color_tmp = [0 for _ in range(n + 1)]
    for x in more_than_one:
        if color_tmp[x] == 0:
            dfs_2(x)
    fill_more_than_one(color_tmp, answer)

    color_tmp = [0 for _ in range(n + 1)]
    for x in in_cycles:
        if color_tmp[x] == 0:
            dfs_2(x)
    fill_inf(color_tmp, answer)

    print(' '.join(map(str, answer[1:])))