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.*;
import static java.lang.Math.*;

public class G implements Runnable{
	static int n;
	static List<Integer>[] adj;
	static List<Integer>[] radj;
	static Stack<Integer> stack;
	static int[] scc;
	static void dfs1(int i) {
		if (scc[i]++!=0) return;
		for (int x: adj[i]) dfs1(x);
		stack.push(i);
	}
	static void dfs2(int i, int v) {
		if (scc[i] != -1) return;
		scc[i]=v;
		for (int x: radj[i]) dfs2(x,v);
	}
	static List<Integer>[] nadj;
	static List<Integer>[] nradj;
	static long[] dp;
	static void dfs3(int i, long mark) {
		// We work with sccs in here. All adj are sccs.
		if (infs[i]) mark = max(mark,(long)(1e12));
		if (dub[i]) mark = max(mark, 2);
		//System.out.println(infs[i] + " Yo " + dp[i]);
		if (dp[i] >= mark) return;
		//System.out.println("hallo " + i);
		dp[i] = max(dp[i],mark);
		//System.out.println("Solution to " + i + ": " + mark);
		for (int x: nadj[i]) {
			//System.out.println("from " + i + " to " + x);
			dfs3(x,dp[i]);
		}
	}
	static boolean[] dub;
	static boolean[] infs;
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		new Thread(null, new G(), "peepee", 1<<28).start();
		
	}
	
	public void run() {
		try {

			int t = readInt();
			while(t-->0) {
				br.readLine();
				n=readInt();
				// Compress all SCCs, and then have a marked value for each of them.
				// Any path visitable from an SCC that was visitable from 1 is infinite.
				// The ending result is a simple DAG. We can do a DP on number of paths
				// let INF be like 1e9, and then add up inf(cap at 1e9).
				adj = new List[n];
				radj = new List[n];
				nadj = new List[n];
				nradj = new List[n];
				boolean[] self = new boolean[n];
				for (int i = 0; i < n; i++) {
					adj[i] = new ArrayList<Integer>();
					radj[i] = new ArrayList<Integer>();
					nadj[i] = new ArrayList<Integer>();
					nradj[i] = new ArrayList<Integer>();
				}
				int m =readInt();
				while(m-->0) {
					int x =readInt()-1;
					int y= readInt()-1;
					if(x==y) self[x]=true;
					else {
						adj[x].add(y);
						radj[y].add(x);
					}
				}
				stack = new Stack<Integer>();
				scc = new int[n];
				dfs1(0);
				boolean[] dubs = new boolean[n];
				for (int i = 0;  i <n; i++) {
					if (scc[i] > 1) dubs[i] = true;
				}
				for (int i = 1; i  <n; i++) dfs1(i);
				Arrays.fill(scc,-1);
				int v= 0;
				infs = new boolean[n];
				dub = new boolean[n];
				while(!stack.isEmpty()) {
					int h = stack.pop();
					if (scc[h] == -1) {
						if(self[h])infs[v]=true;
						if (dubs[h])dub[v]=true;
						dfs2(h,v++);
					}
					
				}
				int[] cnt = new int[n];
				// From this point on, we think about things in terms of their SCC. NOT their actual index.
				// So pretty much always use scc[i] unless I can justify it otherwise.
				for (int i = 0; i < n; i++) {
					if (++cnt[scc[i]]>1) {
						infs[scc[i]]=true;
					}
				}
				for (int i = 0; i < n;i++) {
					for (int x: adj[i]) {
						// Not scc[i] because adj[i] for obv reasons
						if (scc[x] == scc[i]) continue;
						nadj[scc[i]].add(scc[x]);
						nradj[scc[x]].add(scc[i]);
					}
				}
				//for (int i = 0;  i< n; i++) out.print(scc[i] + " ");
				//out.println();
				//for (int i =0; i < n; i++) out.print(b);
				dp = new long[n];
				Arrays.fill(dp, 0);
				dfs3(scc[0],1);
				for (int i = 0; i < n; i++) {
					// This is fine because we're working with scc's, and this makes sense. 
					if (dp[i] > 1e11) dp[i]=-1;
					else dp[i] = min(dp[i],2);
				}
				for (int i = 0; i < n; i++) {
					if (dp[scc[i]]>1e11) dp[scc[i]] = -1;
					out.print(dp[scc[i]] + " ");
				}
				out.println();
			}
			
			out.close();
		}
		catch (IOException e){}
	}
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
	
}