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.io.*;
import java.util.*;

/*


*/

public class D {
	static FastReader sc=null;
	static int cnt[];
	static ArrayList<Integer> adj[];
	
	public static void main(String[] args) {
		sc=new FastReader();
		PrintWriter out=new PrintWriter(System.out);
		int t=sc.nextInt();
		//we have to find a cycle that can be used to get to 'v' -> CYCLE IN DIRECTED GRAPH|TOPSORT
		//we have to find if there exists a path from '1' to 'v' -> DFS
		//we have to find if there exists a duplicate path from '1' to 'v'-> DFS
		while(t-->0) {
			int n=sc.nextInt(),m=sc.nextInt();
			adj=new ArrayList[n];
			cnt=new int[n];
			for(int i=0;i<n;i++)adj[i]=new ArrayList<>();
			
			int indeg[]=new int[n];
			
			for(int i=0;i<m;i++) {
				int v=sc.nextInt()-1,w=sc.nextInt()-1;
				adj[v].add(w);
				indeg[w]++;
			}
			
			dfs(0);
			int ans[]=new int[n];
			for(int i=0;i<n;i++) {
				if(cnt[i]==0) {
					for(int e:adj[i])indeg[e]--;
				}
			}
			ArrayDeque<Integer>topSort=new ArrayDeque<>();
			for(int i=0;i<n;i++)if(cnt[i]>0 && indeg[i]==0)topSort.add(i);
			
			boolean cycle[]=new boolean[n];
			Arrays.fill(cycle, true);
			
			while(!topSort.isEmpty()) {
				int q=topSort.remove();
				cycle[q]=false;
				for(int e:adj[q]) {
					indeg[e]--;
					if(indeg[e]==0 && (cnt[e]>0))topSort.add(e);
				}
			}
			for(int i=0;i<n;i++)if(cnt[i]>0 && cycle[i])ans[i]=-1;
			
			for(int i=0;i<n;i++)if(ans[i]!=-1)ans[i]=cnt[i];
			for(int e:ans)out.print(e+" ");
			out.println();
		}
		out.close();
		
		
		
	}
	static void dfs(int v) {
		if(cnt[v]==2)return;
		cnt[v]++;
		for(int e:adj[v]) {
			dfs(e);
		}
	}
	
	
	static int[] ruffleSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al);
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	static void print(int a[]) {
		for(int e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	static class FastReader{
		
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				}
				catch(IOException e){
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[]=new int[n];
			for(int i=0;i<n;i++)a[i]=sc.nextInt();
			return a;
		}
	}
}