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 <cstdio>
#define print(x,y) write(x),putchar(y)

template <class T>
inline T read(const T sample) {
	T x=0; char s; bool f=0;
	while((s=getchar())>'9' or s<'0')
		f|=(s=='-');
	while(s>='0' and s<='9')
		x=(x<<1)+(x<<3)+(s^48),
		s=getchar();
	return f?-x:x;
}

template <class T>
inline void write(const T x) {
	if(x<0) {
		putchar('-'),write(-x);
		return;
	}
	if(x>9) write(x/10);
	putchar(x%10^48);
}

#include <queue>
#include <vector>
using namespace std;

const int maxn=4e5+5;

struct Kosaraju {
	int n,m,sta[maxn],tp,bl[maxn];
	int siz[maxn];
	bool vis[maxn],cir[maxn];
	vector <int> e0[maxn],e1[maxn];
	
	void dfs1(int u) {
		vis[u]=1;
		for(int i=0;i<e1[u].size();++i)
			if(!vis[e1[u][i]])
				dfs1(e1[u][i]);
		sta[++tp]=u;
	}
	
	void dfs2(int u) {
		bl[u]=tp; vis[u]=1;
		++siz[tp];
		if(cir[u]) siz[tp]=1e9;
		for(int i=0;i<e0[u].size();++i)
			if(!vis[e0[u][i]])
				dfs2(e0[u][i]);
	}
	
	void input() {
		n=read(9),m=read(9);
		for(int i=1;i<=n;++i)
			e0[i].clear(),
			e1[i].clear(),
			vis[i]=cir[i]=0,
			siz[i]=0;
		int u,v; tp=0;
		for(int i=1;i<=m;++i) {
			u=read(9),v=read(9);
			if(u^v)
				e0[u].push_back(v),
				e1[v].push_back(u);
			else cir[u]=1;
		}
		for(int i=1;i<=n;++i)
			if(!vis[i]) dfs1(i);
		for(int i=1;i<=n;++i)
			vis[i]=0;
		tp=0;
		for(int i=n;i>=1;--i)
			if(!vis[sta[i]])
				++tp,dfs2(sta[i]);
	}
} G0;
struct DAG {
	vector <int> e[maxn];
	int in[maxn],ans[maxn],cnt;
	queue <int> q;

	void Link() {
		cnt=G0.tp;
		for(int i=1;i<=cnt;++i)
			in[i]=0,e[i].clear();
		for(int i=1;i<=G0.n;++i)
			for(int j=0;j<G0.e0[i].size();++j)
				if(G0.bl[i]^G0.bl[G0.e0[i][j]])
					e[G0.bl[i]].push_back(G0.bl[G0.e0[i][j]]),
					++in[G0.bl[G0.e0[i][j]]];
	}
	
	int infty(int x) {
		return G0.siz[x]>1?-1:1;
	}
	
	int inc(int x,int y) {
		return x+y>=2?2:x+y;
	}
	
	void topol() {
		for(int i=1;i<=cnt;++i) {
			if(!in[i])	
				q.push(i);
			ans[i]=0;
		}
		ans[G0.bl[1]]=infty(G0.bl[1]);
		while(!q.empty()) {
			int u=q.front(); q.pop();
			for(int i=0;i<e[u].size();++i) {
				int v=e[u][i];
				--in[v];
				if(!in[v]) q.push(v);
				if(ans[u] and infty(v)==-1)
					ans[v]=-1;
				if(v==G0.bl[1] or ans[v]==-1) 
					continue;
				if(ans[u]==-1)
					ans[v]=-1;
				else ans[v]=inc(ans[v],ans[u]);
			}
		}
	}
	
	void output() {
		for(int i=1;i<=G0.n;++i)
			print(ans[G0.bl[i]],' ');
		puts("");
	}
} G;

int main() {
	for(int T=read(9);T;--T) {
		G0.input();
		G.Link();
		G.topol();
		G.output();
	}
	return 0;
}