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 G {

	static ArrayList<ArrayList<Integer>> adjl=new ArrayList<>();
	static int[] indeg;
	static boolean[] vis;
	static int[] ans;
	
	static void dfs(int v,int p) {
		vis[v]=true;
		for(int child:adjl.get(v)) {
			if(!vis[child]) {
				dfs(child,v);
			}
		}
	}
	
	static void multiBFS(LinkedList<Integer> from,int n) {
		boolean[] vis2=new boolean[n+1];
		for(int z:from) {
			ans[z]=-1;
			vis[z]=true;
		}
		while(!from.isEmpty()) {
			int cur=from.removeLast();
			for(int child:adjl.get(cur)) {
				if(!vis2[child]) {
					ans[child]=-1;
					vis2[child]=true;
					from.addLast(child);
				}
			}
		}
	}
	
	static void dfs2(int v,int p) {
		if(ans[v]==2 || ans[v]==-1) {
			return;
		}
		++ans[v];
		for(int child:adjl.get(v)) {
			if(child!=p) {
				dfs2(child,v);
			}
		}
	}
	

	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		Reader sc=new Reader();
		int t=sc.nextInt();
		PrintWriter pw=new PrintWriter(System.out);
		while(t-->0) {
			adjl.clear();
			int n=sc.nextInt();
			int m=sc.nextInt();
			indeg=new int[n+1];
			vis=new boolean[n+1];
			ans=new int[n+1];
			
			for(int i=0;i<=n;i++) {
				adjl.add(new ArrayList<>());
			}
			for(int i=0;i<m;i++) {
				int src=sc.nextInt();
				int dest=sc.nextInt();
				++indeg[dest];
				adjl.get(src).add(dest);
			}
			dfs(1,1);
			for(int i=1;i<=n;i++) {
				if(!vis[i]) {
					for(int child:adjl.get(i)) {
						--indeg[child];
					}
				}
			}

			LinkedList<Integer> q=new LinkedList<>();
			
			for(int i=1;i<=n;i++) {
				if(indeg[i]==0 && vis[i]) {
					q.addLast(i);
				}
			}
			while(!q.isEmpty()) {
				int cur=q.removeFirst();
				for(int child:adjl.get(cur)) {
					--indeg[child];
					if(indeg[child]==0) {
						q.addLast(child);
					}
				}
			}
			LinkedList<Integer> cyclestarts=new LinkedList<>();
			for(int i=1;i<=n;i++) {
				if(indeg[i]!=0 && vis[i]) {
					cyclestarts.add(i);
				}
			}
			multiBFS(cyclestarts,n);
			dfs2(1,1);
			for(int i=1;i<=n;i++) {
				pw.print(ans[i]+" ");
			}
			pw.println();
		}
		pw.flush();
		sc.close();
	}
	
	static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }

}