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 Main{
	static boolean[]hasCycle;
	static int[]v;
	static int unVis=0,seen=1,finished=2;
	static LinkedList<Integer>[]adj;
	static void dfs(int i) {
		v[i]=seen;
		for(int j:adj[i]) {
			if(v[j]==unVis) {
				dfs(j);
				continue;
			}
			if(v[j]==finished)continue;
			hasCycle[j]=true;
		}
		v[i]=finished;
	}
	static boolean[][]vis;
	static void dfs(int i,int pathes) {
		vis[i][pathes]=true;
		for(int j:adj[i]) {
			if((pathes==2 || hasCycle[j])) {
				if(!vis[j][2])
					dfs(j, 2);
				continue;
			}
			if(pathes==1 || vis[j][0]) {
				if(!vis[j][1]) {
					dfs(j, 1);
				}
				continue;
			}
			if(!vis[j][0]) {
				dfs(j, 0);
			}
		}
		
	}
	static void main() throws Exception{
		int n=sc.nextInt(),m=sc.nextInt();
		adj=new LinkedList[n];
		for(int i=0;i<n;i++)adj[i]=new LinkedList<Integer>();
		for(int i=0;i<m;i++)adj[sc.nextInt()-1].add(sc.nextInt()-1);
		hasCycle=new boolean[n];
		v=new int[n];
		vis=new boolean[n][3];
		dfs(0);
		dfs(0, hasCycle[0]?2:0);
		for(int i=0;i<n;i++) {
			if(vis[i][2]) {
				pw.print(-1+" ");
				continue;
			}
			if(vis[i][1]) {
				pw.print(2+" ");
				continue;
			}
			if(vis[i][0]) {
				pw.print(1+" ");
				continue;
			}
			pw.print(0+" ");
		}
		pw.println();
	}
	public static void main(String[] args) throws Exception{
		sc=new MScanner(System.in);
		pw = new PrintWriter(System.out);
		int tc=1;
        tc=sc.nextInt();
		for(int i=1;i<=tc;i++) {
//			pw.printf("Case #%d: ", i);
			main();
		}
		
		pw.flush();
	}
	static PrintWriter pw;
	static MScanner sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
	 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
	 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
	 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
	 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
	 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	 
		public boolean ready() throws IOException {
			return br.ready();
		}
	 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
		
	}
	static void dbg(int[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void dbg(long[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void sort(int[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void sort(long[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}