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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Solution {
    static int M = 1_000_000_007;
    static Random rng = new Random();

    private static int[] testCase(int n, int m, int[][] ab) {
        int[] ans = new int[n];
        ArrayList<Integer>[] adjList = new ArrayList[n];
        int[] visited = new int[n];

        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
        }

        for (int[] abi : ab) {
            adjList[abi[0] - 1].add(abi[1] - 1);
        }

        dfs(adjList, 0, visited, ans);

        return ans;
    }

    private static void dfs(ArrayList<Integer>[] adjList, int curr, int[] visited, int[] ans) {
        if (ans[curr] == -1) {
            return;
        }

        if (visited[curr] == 2) {
            updateFinitePaths(adjList, curr, ans);
        } else if (visited[curr] == 1) {
            updateInfinitePaths(adjList, curr, ans);
        } else {
            visited[curr] = 1;
            ans[curr] = 1;

            for (int v : adjList[curr]) {
                dfs(adjList, v, visited, ans);
            }

            visited[curr] = 2;
        }
    }

    private static void updateFinitePaths(ArrayList<Integer>[] adjList, int curr, int[] ans) {
        if (ans[curr] == -1 || ans[curr] == 2) {
            return;
        }

        ans[curr] = 2;

        for (int v : adjList[curr]) {
            updateFinitePaths(adjList, v, ans);
        }
    }

    private static void updateInfinitePaths(ArrayList<Integer>[] adjList, int curr, int[] ans) {
        if (ans[curr] == -1) {
            return;
        }

        ans[curr] = -1;

        for (int v : adjList[curr]) {
            updateInfinitePaths(adjList, v, ans);
        }
    }


    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();  // Scanner has functions to read ints, longs, strings, chars, etc.
        //in.nextLine();
        for (int tt = 1; tt <= t; ++tt) {
            int n = in.nextInt(), m = in.nextInt();
            int[][] ab = new int[m][2];

            for (int i = 0; i < m; i++) {
                ab[i] = in.readArray(2);
            }

            int[] ans = testCase(n, m, ab);

            for (int an : ans) {
                out.print(an);
                out.print(' ');
            }
            out.println();
        }

        out.close();
    }

    private static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        boolean hasNext() {
            return st.hasMoreTokens();
        }

        char[] readCharArray(int n) {
            char[] arr = new char[n];
            try {
                br.read(arr);
                br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return arr;
        }

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }

    private static void sort(int[] arr) {
        int temp, idx;

        for (int i = arr.length - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = arr[i];
            arr[i] = arr[idx];
            arr[idx] = temp;
        }

        Arrays.sort(arr);
    }

    private static void sort(long[] arr) {
        long temp;
        int idx;

        for (int i = arr.length - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = arr[i];
            arr[i] = arr[idx];
            arr[idx] = temp;
        }

        Arrays.sort(arr);
    }

    private static <T> void sort(T[] arr) {
        T temp;
        int idx;

        for (int i = arr.length - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = arr[i];
            arr[i] = arr[idx];
            arr[idx] = temp;
        }

        Arrays.sort(arr);
    }

    private static <T> void sort(T[] arr, Comparator<? super T> cmp) {
        T temp;
        int idx;

        for (int i = arr.length - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = arr[i];
            arr[i] = arr[idx];
            arr[idx] = temp;
        }

        Arrays.sort(arr, cmp);
    }

    private static <T extends Comparable<? super T>> void sort(ArrayList<T> list) {
        T temp;
        int idx;

        for (int i = list.size() - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = list.get(i);
            list.set(i, list.get(idx));
            list.set(idx, temp);
        }

        Collections.sort(list);
    }

    private static <T> void sort(ArrayList<T> list, Comparator<? super T> cmp) {
        T temp;
        int idx;

        for (int i = list.size() - 1; i > 0; i--) {
            idx = rng.nextInt(i + 1);

            temp = list.get(i);
            list.set(i, list.get(idx));
            list.set(idx, temp);
        }

        Collections.sort(list, cmp);
    }
}