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.util.*;

public class Solution{
    static ArrayList<Integer>[] gr;
    static Set<Integer>[] s;
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        StringBuilder sb=new StringBuilder();
        while(t-->0){
            int n=sc.nextInt(), m=sc.nextInt();
            gr=new ArrayList[n+1];
            s=new HashSet[2];
            for(int i=0;i<=n;i++){
                if(i<2) s[i]=new HashSet<Integer>();
                gr[i]=new ArrayList<Integer>();
            }
            int[] col=new int[n+1];
            for(int i=0;i<m;i++){
                int a=sc.nextInt(),b=sc.nextInt();
                gr[a].add(b);
            }
            dfs(1,col,true);
            int[][] c=new int[2][n+1];
            for(int i=0;i<2;i++){
                for(int a:s[i])dfs(a,c[i],false);
            }
            
            for(int i=1;i<=n;i++){
                int color=0;
                if(col[i]!=0){
                    if(c[0][i]>0)color=-1;
                    else if(c[1][i]>0)color=2;
                    else color=1;
                }
                sb.append(color+" ");
            }
            sb.append("\n");
            
        }
        System.out.println(sb);
    }
    static void dfs(int n,int[] col,boolean ok){
        col[n]=1;
        for(int child:gr[n]){
            if(col[child]==0)dfs(child,col,ok);
            else if(ok)s[col[child]-1].add(child);
        }
        col[n]=2;
    }
}