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 contestA {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-->0){
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            ArrayList<Integer>[] g = new ArrayList[n];
            for(int i=0;i<n;++i) g[i] = new ArrayList<>();
            while(m-->0){
                int u = scanner.nextInt() - 1;
                int v = scanner.nextInt() - 1;
                g[u].add(v);
            }
            int[] mk = new int[n];
            ArrayList<Integer> s1 = new ArrayList<>();
            ArrayList<Integer> s2 = new ArrayList<>();
            dfs(0,mk,g,s1,s2);
            int[] c1 = new int[n];
            int[] c2 = new int[n];
            for(int u:s1) dfs(u,c1,g);
            for(int u:s2) dfs(u,c2,g);
            for(int i=0;i<n;++i){
                if(mk[i]==0) System.out.print("0 ");
                else if(c1[i]==1) System.out.print("-1 ");
                else if(c2[i]==1) System.out.print("2 ");
                else System.out.print("1 ");
            }
            System.out.println();
        }
    }
    static void dfs(int u,int[] mk,ArrayList<Integer>[]g,ArrayList<Integer> s1,ArrayList<Integer> s2){
        mk[u] = 1;
        for(int v:g[u]){
            if(mk[v]==0) dfs(v,mk,g,s1,s2);
            else if(mk[v]==1) s1.add(v);
            else s2.add(v);
        }
        mk[u] = 2;
    }
    static void dfs(int u,int[] mk,ArrayList<Integer>[]g){
        mk[u] = 1;
        for(int v:g[u]){
            if(mk[v]==0) dfs(v,mk,g);
        }
    }
}