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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            bu.readLine();
            String s[]=bu.readLine().split(" ");
            int n=Integer.parseInt(s[0]),m=Integer.parseInt(s[1]);
            ArrayList<Integer> g[]=new ArrayList[n];

            int i;
            for(i=0;i<n;i++) g[i]=new ArrayList<>();
            for(i=0;i<m;i++)
            {
                s=bu.readLine().split(" ");
                int u=Integer.parseInt(s[0])-1,v=Integer.parseInt(s[1])-1;
                g[u].add(v);
            }

            int ans[]=new int[n],vis[]=new int[n];
            ans[0]=1;
            dfs(g,ans,vis,0);
            for(i=0;i<n;i++) sb.append(ans[i]+" ");
            sb.append("\n");
        }
        System.out.print(sb);
    }

    static void dfs(ArrayList<Integer> g[],int ans[],int vis[],int n)
    {
        vis[n]=1;
        for(int x:g[n])
        if(ans[n]==-1)
        {
            if(ans[x]!=-1)
            {
                ans[x]=-1;
                dfs(g,ans,vis,x);
            }
        }
        else
        {
            if(vis[x]==1)
            {
                ans[x]=-1;
                dfs(g,ans,vis,x);
            }
            else if(vis[x]==0)
            {
                ans[x]=ans[n];
                dfs(g,ans,vis,x);
            }
            else if(ans[x]==1)
            {
                ans[x]=2;
                dfs(g,ans,vis,x);
            }
        }
        vis[n]=2;
    }
}