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:
// Problem: G. How Many Paths?
// Contest: Codeforces - Codeforces Round #731 (Div. 3)
// URL: https://codeforces.com/contest/1547/problem/G
// Memory Limit: 512 MB
// Time Limit: 4000 ms

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define ll          long long
#define pb          push_back
#define endl        '\n'
#define mod         (ll)1000000007
using namespace std;
ll power(ll a , ll b,ll modi)
{
    a%=modi;
    ll res = 1 ;
    while(b)
    {
        if(b%2) {
            res = (res * a) % modi ;
        }
        b/=2 ;
        a = (a*a) % modi ;
    }
    return res ;
}
//unordered_map hash function faster
// struct custom_hash {
    // static uint64_t splitmix64(uint64_t x) {
        // // http://xorshift.di.unimi.it/splitmix64.c
        // x += 0x9e3779b97f4a7c15;
        // x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        // x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        // return x ^ (x >> 31);
    // }
//  
    // size_t operator()(uint64_t x) const {
        // static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        // return splitmix64(x + FIXED_RANDOM);
    // }
// };
//ll fac[200001];
//void facto(){
//	fac[0]=1;
//	for(ll i=1;i<=200000;i++)
//	fac[i]=(fac[i-1]*i)%mod;
//}
//ll ncr(ll n,ll r){
//	ll num=fac[n];
//	num=(num*power(fac[n-r],mod-2,mod))%mod;
//	num=(num*power(fac[r],mod-2,mod))%mod;
//	return num;
//}
vector<int>adj[400001],adj1[400001];
bool mark[400001];
int ans[400001];
int incoming[400001];
void dfs(int s){
	mark[s]=1;
	for(int j=0;j<(int)adj[s].size();j++)
	if(!mark[adj[s][j]])
	dfs(adj[s][j]);
}
void bfs(){
	queue<int>q;
    if(incoming[1]==0)
    q.push(1);
    while(!q.empty()){
    	int f=q.front();
    	q.pop();
    	ans[f]=0;
    	for(int i=0;i<(int)adj1[f].size();i++){
    		incoming[adj1[f][i]]--;
    		if(!incoming[adj1[f][i]])
    		q.push(adj1[f][i]);
    	}
    }
}
void dfs1(int s){
	for(int i=0;i<(int)adj1[s].size();i++){
		if(ans[adj1[s][i]]==-1)
		continue;
		if(ans[adj1[s][i]]<2){
			ans[adj1[s][i]]++;
			dfs1(adj1[s][i]);
		}
	}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
/*#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
 #endif*/
int t;
	cin>>t;
	while(t--)
	{
	    int n,m;
	    cin>>n>>m;
	    for(int i=1;i<=n;i++){
		    adj[i].clear();
		    adj1[i].clear();
		    ans[i]=-1;
		    incoming[i]=0;
		    mark[i]=0;
	    }
	    for(int i=0;i<m;i++){
	    	int u,v;
	    	cin>>u>>v;
	    	adj[u].pb(v);
	    }
	    dfs(1);
	    for(int i=1;i<=n;i++){
	    	if(mark[i]){
	    		for(int j=0;j<(int)adj[i].size();j++){
	    			if(mark[adj[i][j]]){
		    			adj1[i].pb(adj[i][j]);
		    			incoming[adj[i][j]]++;
	    			}
	    		}
	    	}
	    	else
	    	ans[i]=0;
	    }
	    bfs();
	    if(ans[1]!=-1){
	        dfs1(1);
	        ans[1]=1;
	    }
	    for(int i=1;i<=n;i++)
	    cout<<ans[i]<<" ";
	    cout<<endl;
	}
return 0;
}