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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Solution {

    private static HashSet<Integer> infinite;
    private static HashSet<Integer> finite;
    private static List<Integer> graph[];

    public static void main(String[] args) throws IOException {

        Scanner in=new Scanner(System.in);
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringBuffer out = new StringBuffer();

        int T = in.nextInt();

        OUTER:
        while(T-->0) {

            int n=in.nextInt(), m=in.nextInt();

            graph=new LinkedList[n];
            for(int i=0; i<n; i++)
                graph[i]=new LinkedList<>();

            for(int i=0; i<m; i++) {
                graph[in.nextInt()-1].add(in.nextInt()-1);
            }

            int path[]=new int[n];
            infinite=new HashSet<>();
            finite=new HashSet<>();
            dfs(0, path, new HashSet<>());

            for(int item: finite) {
                visit(item, path, 2);
            }

            for(int item: infinite) {
                visit(item, path, -1);
            }

            for(int i=0; i<n; i++) {
                if(path[i]>1) {
                    out.append("2 ");
                } else {
                    out.append(path[i]+" ");
                }
            }
            out.append("\n");
        }

        System.out.print(out);
    }

    private static void visit(int vertex, int path[], int value) {
        if(path[vertex]==value)
            return;
        path[vertex]=value;
        for(int item: graph[vertex])
            visit(item, path, value);
    }

    private static int last=-1;

    private static void dfs(int vertex, int path[], HashSet<Integer> visited) {
        if(path[vertex]==1) {
            finite.add(vertex);
            return;
        }

        path[vertex]=1;
        visited.add(vertex);
        for(int item: graph[vertex]) {
            if(visited.contains(item)) {
                last=item;
                break;
            } else {
                dfs(item, path, visited);
                if(last!=-1) {
                    break;
                }
            }
        }
        visited.remove(vertex);

        if(last!=-1) {
            infinite.add(vertex);
        }
        if(last==vertex) {
            last=-1;
        }
    }

    private static int gcd(int a, int b) {
        if (a==0)
            return b;
        return gcd(b%a, a);
    }

    private static int toInt(String s) {
        return Integer.parseInt(s);
    }

    private static long toLong(String s) {
        return Long.parseLong(s);
    }
}