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.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStream;

/** Built using CHelper plug-in Actual solution is at the top */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    FastScanner in = new FastScanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    GHowManyPaths solver = new GHowManyPaths();
    int testCount = Integer.parseInt(in.next());
    for (int i = 1; i <= testCount; i++) solver.solve(i, in, out);
    out.close();
  }

  static class GHowManyPaths {
    public void solve(int testNumber, FastScanner in, PrintWriter out) {
      int n = in.nextInt();
      int m = in.nextInt();
      List<Integer>[] g = GraphUtils.nextD(in, n, m, true);
      int[] clus = new SCCNoRec(g).getClus();
      List<Integer>[] cg = GraphUtils.condense(g, clus);
      int clusSize = cg.length;
      int[] nums = new int[clusSize];
      for (int cur = 0; cur < n; ++cur) {
        for (Integer next : g[cur]) {
          int fc = clus[cur];
          int tc = clus[next];
          if (fc == tc) ++nums[fc];
        }
      }

      int[] dp = new int[clusSize];
      dp[clus[0]] = 1;
      for (int i = clusSize - 1; i >= 0; --i) {
        if (nums[i] > 0 && dp[i] != 0) {
          dp[i] = -1;
        }
        for (Integer next : cg[i]) {
          if (dp[i] == -1 || dp[next] == -1) {
            dp[next] = -1;
          } else {
            dp[next] += dp[i];
            dp[next] = Math.min(2, dp[next]);
          }
        }
      }

      for (int i = 0; i < n; ++i) {
        out.print(dp[clus[i]] + " ");
      }
      out.println();
    }
  }

  static class FastScanner implements PReader {
    BufferedReader reader;
    StringTokenizer tokenizer;

    public FastScanner(InputStream inputStream) {
      reader = new BufferedReader(new InputStreamReader(inputStream), 32768);
      tokenizer = null;
    }

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

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

  static class SCCNoRec {
    List<Integer>[] g;
    int[] clus;
    int n;

    public SCCNoRec(List<Integer>[] g) {
      this.g = g;
      this.n = g.length;
      clus = new int[n];
      decomposeToSCCNoRec();
    }

    public void decomposeToSCCNoRec() {
      int n = g.length;
      int[] stack = new int[n + 1];
      int[] ind = new int[n + 1];
      int[] ord = new int[n];
      Arrays.fill(ord, -1);
      int[] low = new int[n];
      Arrays.fill(low, -1);
      int sp = 0;
      int id = 0; // preorder
      int cid = 0;
      int[] cstack = new int[n + 1];
      int csp = 0;
      boolean[] incstack = new boolean[n];
      for (int i = 0; i < n; i++) {
        if (ord[i] == -1) {
          ind[sp] = 0;
          cstack[csp++] = i;
          stack[sp++] = i;
          incstack[i] = true;
          while (sp > 0) {
            int cur = stack[sp - 1];
            if (ind[sp - 1] == 0) {
              ord[cur] = low[cur] = id++;
            }
            if (ind[sp - 1] < g[cur].size()) {
              int nex = g[cur].get(ind[sp - 1]);
              if (ord[nex] == -1) {
                ind[sp - 1]++;
                ind[sp] = 0;
                incstack[nex] = true;
                cstack[csp++] = nex;
                stack[sp++] = nex;
              } else {
                // shortcut
                // U.tr(cur, nex, incstack[nex], low[nex], stack);
                if (incstack[nex])
                  low[cur] = Math.min(low[cur], ord[nex]); // low is ok too, but slower.
                ind[sp - 1]++;
              }
            } else {
              if (ord[cur] == low[cur]) {
                while (csp > 0) {
                  incstack[cstack[csp - 1]] = false;
                  clus[cstack[--csp]] = cid;
                  if (cstack[csp] == cur) break;
                }
                cid++;
              }
              if (--sp >= 1) low[stack[sp - 1]] = Math.min(low[stack[sp - 1]], low[stack[sp]]);
            }
          }
        }
      }
    }

    public int[] getClus() {
      return clus;
    }
  }

  static class GraphUtils {
    public static List<Integer>[] nextD(PReader in, int n, int m, boolean oneIndexed) {
      int diff = oneIndexed ? 1 : 0;
      List<Integer>[] g = new List[n];
      for (int i = 0; i < n; ++i) g[i] = new ArrayList<>();
      for (int i = 0; i < m; ++i) {
        int u = in.nextInt() - diff;
        int v = in.nextInt() - diff;
        g[u].add(v);
      }
      return g;
    }

    public static List<Integer>[] condense(List<Integer>[] graph, int[] cluster) {
      int n = graph.length;
      int m = 0;
      for (int i = 0; i < n; i++) m = Math.max(m, cluster[i]);
      m++;

      List<Integer>[] condensedGraph = new List[m];
      for (int i = 0; i < m; ++i) {
        condensedGraph[i] = new ArrayList<>();
      }

      for (int i = 0; i < n; i++) {
        for (int j = 0; j < graph[i].size(); j++) {
          condensedGraph[cluster[i]].add(cluster[graph[i].get(j)]);
        }
      }

      for (int i = 0; i < m; i++) {
        List<Integer> cur = condensedGraph[i];
        Collections.sort(cur);
        List<Integer> temp = new ArrayList<>();
        for (int j = 0; j < condensedGraph[i].size(); j++) {
          if ((j == 0 || !cur.get(j).equals(cur.get(j - 1))) && cur.get(j) != i) {
            temp.add(cur.get(j));
          }
        }
        condensedGraph[i] = temp;
      }
      return condensedGraph;
    }
  }

  static interface PReader {
    int nextInt();
  }
}