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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Omar Khair
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        GHowManyPaths solver = new GHowManyPaths();
        solver.solve(1, in, out);
        out.close();
    }

    static class GHowManyPaths {
        static ArrayList<Integer>[] graph;
        static int n;
        static int m;
        static int[] dfsNum;
        static boolean[] is2path;
        static boolean[] isInfinitePath;
        static final int UNVISITED = 0;
        static final int VISITED = 1;
        static final int EXPLORED = 2;

        public void solve(int testNumber, Scanner sc, PrintWriter out) {
            int tt = sc.nextInt();
            while (tt-- > 0) {
                n = sc.nextInt();
                m = sc.nextInt();
                graph = new ArrayList[n];
                for (int i = 0; i < n; i++)
                    graph[i] = new ArrayList<>();
                while (m-- > 0) {
                    int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
                    graph[u].add(v);
                }
                dfsNum = new int[n];
                is2path = new boolean[n];
                isInfinitePath = new boolean[n];
                edgeTypeDFS(0);
                for (int i = 0; i < n; i++) {
                    if (is2path[i])
                        reachabilityDFS(i, is2path);
                    if (isInfinitePath[i])
                        reachabilityDFS(i, isInfinitePath);
                }
                for (int i = 0; i < n; i++) {
                    int ans = 0;
                    if (dfsNum[i] == UNVISITED)
                        ans = 0;
                    if (dfsNum[i] == VISITED)
                        ans = 1;
                    if (is2path[i])
                        ans = 2;
                    if (isInfinitePath[i])
                        ans = -1;
                    out.print(ans + " ");

                }
                out.println();
            }
        }

        private void reachabilityDFS(int u, boolean[] vis) {
            for (int v : graph[u])
                if (!vis[v]) {
                    vis[v] = true;
                    reachabilityDFS(v, vis);
                }
        }

        private void edgeTypeDFS(int u) {
            dfsNum[u] = EXPLORED;
            for (int v : graph[u]) {
                if (dfsNum[v] == UNVISITED)
                    edgeTypeDFS(v);
                else if (dfsNum[v] == EXPLORED) {
                    // this is a cycle
                    isInfinitePath[u] = true;
                } else if (dfsNum[v] == VISITED) {
                    // this is a multipath
                    is2path[v] = true;
                }
            }
            dfsNum[u] = VISITED;
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}