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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define all(v) v.begin(),v.end()
#define mod 1000000007 
#define N 500010
#define inf LLONG_MAX

//---------------------------------------------------------------
vector<int> graph[N];
vector<int> incoming[N];
vector<int> visited(N,false);
vector<int> ans(N,0);
map<int,bool> mp;
vector<int> special;
vector<int> colour(N,0);
vector<int> parent(N,-1);
vector<bool> store(N,false);

//---------------------------------------------------------------

void dfs(int node,int mark)
{
	visited[node]=true;
	ans[node]=mark;
	
	for(auto x:graph[node])
	{
		if(!visited[x])
		{
			dfs(x,mark);
		}
	}
}


void dfs2(int node)
{
	visited[node]=true;
	if(ans[node]==1)
	{
		ans[node]=2;
	}
	for(auto x:graph[node])
	{
		if(!visited[x])
		{
			dfs2(x);
		}	
	}
}


//---------------------------------------------------------------

void cycle_wala_dfs(int node)
{
	visited[node]=true;
	colour[node]=1;
	for(auto x:graph[node])
	{
		if(!visited[x])
		{
			parent[x]=node;
			cycle_wala_dfs(x);
		}
		else if(visited[x] and colour[x]==1)
		{
			//part of cycle 
			int curr=node;
			while(curr!=x)
			{
				if(ans[curr]==-1)
				{
					break;
				}
				ans[curr]=-1;
				special.push_back(curr);
				curr=parent[curr];
			}
			ans[x]=-1;
		}
	}
	colour[node]=2;
}


//---------------------------------------------------------------

void clr_vis(int n)
{
	for(int i=0;i<n;i++)
    {
    	visited[i]=false;
    }
        
}


void clr(int n)
{
	for(int i=0;i<=n;i++)
	{
		graph[i].clear();
		incoming[i].clear();
		visited[i]=false;
		ans[i]=0;
		store[i]=false;
		parent[i]=-1;
		colour[i]=0;
		
	}
	mp.clear();
	special.clear();
}

signed main()
{
    fast
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        clr(n);
        
        
        for(int i=0;i<m;i++)
        {
            int u,v;
            cin>>u>>v;
            u--;v--;
            
            if(u==v)
            {
            	mp[u]=true;
            }
            else 
            {
            	graph[u].push_back(v);
            	incoming[v].push_back(u);
            }
            
        }
        dfs(0,1);
        for(auto x:mp)
        {
        	if(visited[x.first])
        	{
        		special.push_back(x.first);
        	}
        }
        for(int i=0;i<n;i++)
        {
        	store[i]=visited[i];
        }
        clr_vis(n);
        
        cycle_wala_dfs(0);
        
        clr_vis(n);
        
        for(auto x:special)
        {
        	if(!visited[x])
        		dfs(x,-1);
        }
        
        
        vector<int> multiple;
        for(int i=0;i<n;i++)
        {
        	if(ans[i]==1)
        	{
        		int cnt=0;
        		for(auto x:incoming[i])
        		{
        			if(store[x])
        			{
        				cnt++;
        			}
        		}
        		if(cnt>1)
        		{
        			multiple.push_back(i);
        		}
        	}
        }
        
        clr_vis(n);
        
        for(auto x:multiple)
        {
        	if(!visited[x])
        	{
        		dfs2(x);
        	}
        }
        
        
        
        // printing the final ans 
        for(int i=0;i<n;i++)
        {
        	cout<<ans[i]<<" ";
        }
        cout<<"\n";
    }
}

//Note : Always check for the following things
//1. Max value of N
//2. Mod value
//3. Test for your own case
//4. Check for special values (1 in gcd, corner case etc...)