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.*;
import java.io.*;
 
public class Sol{
//  ************     n==1  ***************** important test case
 
static int timer;
static ArrayList<Integer> g[],gt[];
static boolean vis[],connected[],ass[];
static Pair p[];
static int value[];
 
public static void main(String []args){

int times=ni();
while(times-->0){
  int n=ni();int m=ni();
  precomp(n,m);
 
  for(int i=1;i<=n;i++){if(!vis[i])dfs(i,vis);}
  
  Arrays.sort(p);
  
  int i=n;
  while(i>=1){
     int u=p[i].id;
     ArrayList<Integer> al=new ArrayList<>();
     if(ass[u])i--;
     else{Dfs(u,al);
          if(al.size()>1){for(int j=0;j<al.size();j++){value[al.get(j)]=-1;}}
     }
  }
  
  dfs(1,connected);for(i=1;i<=n;i++)vis[i]=false;
  for(i=1;i<=n;i++){if(connected[i]&&value[i]==-1){DFS(i);}
                    else{value[i]=0;}
                    vis[i]=false;}
  
  if(value[1]!=-1){
   dfs1(1,0);
  }
   
  for(i=1;i<=n;i++)out.print(value[i]+" ");
  out.println();

}out.close();}
 
static void dfs(int u,boolean vis[]){
   vis[u]=true;timer++;
   for(int i:g[u]){
     if(!vis[i])dfs(i,vis);
   }
   timer++;
   p[u].out=timer;   
 }
 
static void Dfs(int u,ArrayList<Integer> al){
     ass[u]=true;al.add(u);
     for(int i:gt[u]){
       if(!ass[i])Dfs(i,al);
    }
 }
 
static void DFS(int u){
   value[u]=-1;vis[u]=true;
   for(int i:g[u]){
      if(vis[i])continue;
      DFS(i);   
   }
 }
static void dfs1(int u,int p){
   if(count(u))value[u]=2;
   else value[u]=1;
   vis[u]=true;
   for(int i:g[u]){if(vis[i]||value[i]==-1)continue;dfs1(i,u);}
 }

static boolean count(int u){
   int ans=0;
   for(int i:gt[u]){
      if(connected[i])ans+=Math.max(value[i],1);
   }
   return ans>=2;
 }

static void precomp(int n,int m){
   
  g=new ArrayList[n+1];
  gt=new ArrayList[n+1];
  vis=new boolean[n+1];
  ass=new boolean[n+1];
  connected=new boolean[n+1];
  p=new Pair[n+1];timer=0;
  value=new int[n+1];
 
  for(int i=0;i<=n;i++){p[i]=new Pair(i,0);}
  for(int i=0;i<=n;i++){g[i]=new ArrayList<Integer>();gt[i]=new ArrayList<Integer>();}
  for(int i=1;i<=m;i++){int s=ni();int e=ni();g[s].add(e);gt[e].add(s);if(e==s)value[e]=-1;}
 }

//-----------------Utility--------------------------------------------
 
static int gcd(int a,int b){if(b==0)return a; return gcd(b,a%b);}
 
static int Max=Integer.MAX_VALUE; static long mod=998244353;
 
static int v(char c){return (int)(c-'a')+1;}
 
public static long power(long x, long y )
    {
        //0^0 = 1
        long res = 1L;
        x = x%mod;
        while(y > 0)
        {
            if((y&1)==1)
                res = (res*x)%mod;
            y >>= 1;
            x = (x*x)%mod;
        }
        return res;
    }
 
static class Pair implements Comparable<Pair>{
        int id;int out;
 
        public Pair(int id,int out) {
         
          this.id=id;this.out=out;
        }
        @Override
        public int compareTo(Pair p){return Long.compare(out,p.out);}
        //public int compareTo(Pair p){return id-p.id;}
       }
 
//----------------------I/O---------------------------------------------
 
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static FastReader in=new FastReader(inputStream);
static PrintWriter out=new PrintWriter(outputStream);
 
static class FastReader 
{ 
  BufferedReader br; 
  StringTokenizer st; 
 
        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }
 
  public String next() 
  { 
      while (st == null || !st.hasMoreElements()) 
      { 
          try
          { 
              st = new StringTokenizer(br.readLine()); 
          } 
          catch (IOException  e) 
          { 
              e.printStackTrace(); 
          } 
      } 
      return st.nextToken(); 
  } 
 
  public int nextInt() 
  { 
      return Integer.parseInt(next()); 
  } 
 
  public long nextLong() 
  { 
      return Long.parseLong(next()); 
  } 
 
 public  double nextDouble() 
  { 
      return Double.parseDouble(next()); 
  } 
 
 
  String nextLine() 
  { 
      String str = ""; 
      try
      { 
          str = br.readLine(); 
      } 
      catch (IOException e) 
      { 
          e.printStackTrace(); 
      } 
      return str; 
  } 
}
static int ni() {
        try {
            boolean in = false;
            int res = 0;
            for (;;) {
                int b = System.in.read() - '0';
                if (b >= 0) {
                    in = true;
                    res = 10 * res + b;
                }
                else if (in)
                    return res;
            }
        } catch (IOException e) {
            throw new Error(e);
        }
    }
static long nl(){return in.nextLong();}
static String ns(){return in.nextLine();}
 
}