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>
#define log(a) cerr<<"\033[32m[DEBUG] "<<#a<<'='<<(a)<<" @ line "<<__LINE__<<"\033[0m"<<endl
#define LL long long
#define SZ(x) ((int)x.size()-1)
#define ms(a,b) memset(a,b,sizeof a)
#define F(i,a,b) for(int i=(a);i<=(b);++i)
#define DF(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
inline int read(){char ch=getchar(); int w=1,c=0;
	for(;!isdigit(ch);ch=getchar()) if (ch=='-') w=-1;
	for(;isdigit(ch);ch=getchar()) c=(c<<1)+(c<<3)+(ch^48);
	return w*c;
}
const int N=4e5+10;
int _,n,m,scccnt,dfn[N],low[N],dfscnt,sccnum[N],sccsize[N],f[N],/*du[N],*/sk[N],stop;
bool h[N];
vector<int>v[N],sccv[N];
void tarjan(int now){
	dfn[now]=low[now]=++dfscnt;
	sk[stop++]=now;
	for(int i:v[now]){
		if(!dfn[i]){
			tarjan(i);
			low[now]=min(low[now],low[i]);
		}else if(!sccnum[i]){
			low[now]=min(low[now],dfn[i]);
		}
	}
	if(dfn[now]==low[now]){
		scccnt++;
		do{
			sccnum[sk[--stop]]=scccnt;
            sccsize[scccnt]++;
		}while(sk[stop]!=now);
	}
}
queue<int>q;
signed main(){
	_=read();while(_--){
		n=read(),m=read();
		stop=scccnt=dfscnt=0;
		F(i,1,n)v[i].clear(),sccv[i].clear(),dfn[i]=low[i]=sccnum[i]=sccsize[i]=f[i]=/*du[i]=*/0,h[i]=false;
		F(i,1,m){
			int x=read(),y=read();
			if(x==y)h[x]=true;
			else v[x].push_back(y);
		}
		F(i,1,n)
			if(!dfn[i])tarjan(i);
		// F(i,1,n){
		// 	log(i);
		// 	log(sccnum[i]);
		// }
		F(i,1,n)
			for(int j:v[i])
				if(sccnum[i]!=sccnum[j]&&sccnum[j]!=sccnum[1]){
					sccv[sccnum[i]].push_back(sccnum[j]);
					// log(sccnum[i]);log(sccnum[j]);
					// du[sccnum[j]]++;
				}
		F(i,1,n)
			if(h[i])sccsize[sccnum[i]]++;
		if(sccsize[sccnum[1]]>1)f[sccnum[1]]=-1;
		else f[sccnum[1]]=1;
		q.push(sccnum[1]);
		while(q.size()){
			int x=q.front();q.pop();
			// log(x);
			for(int i:sccv[x]){
				// cout<<i<<endl;
				if(!f[i]){
					f[i]=f[x];
					if(sccsize[i]>1)f[i]=-1;
					q.push(i);
				}else{
					if(f[i]==1&&(f[x]==1||f[x]==2))f[i]=2,q.push(i);
					if((f[i]==1||f[i]==2)&&f[x]==-1)f[i]=-1,q.push(i);
				}
				// if(--du[i]==0)q.push(i);
			}
		}
		F(i,1,n)cout<<f[sccnum[i]]<<(i==n?'\n':' ');
	}
	return 0;
}