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:
// Licykoc yyds!!! &.*^*.&
%:include"bits/stdc++.h"
using namespace std;
#define N (int)(4e5+10) 
vector<int>g1[N],g2[N],g3[N];
stack<int>s;
queue<int>q;
int T,n,m;
int dp[N],len,color[N],ru[N],siz[N],f[N];
bool vis[N],b[N],c[N];
inline void dfs1(int u) {
	vis[u]=true;
	for(int i=0;i<g1[u].size();i++) {
		int v=g1[u][i];
		if (!vis[v]) dfs1(v); 
	} 
	s.push(u);
} 
inline void dfs2(int u) {
	color[u]=len; vis[u]=false; siz[len]++;
	for(int i=0;i<g2[u].size();i++) {
		int v=g2[u][i];
		if (vis[v]) dfs2(v);
	}
}
signed main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin>>T;
	while (T-->0) {
		cin>>n>>m;
		for(int i=1;i<=n;i++) g1[i].clear(),g2[i].clear(),g3[i].clear();
		len=0;
		memset(siz,0,sizeof(siz));
		memset(b,false,sizeof(b));
		memset(c,false,sizeof(c));
		for(int i=1;i<=n;i++) f[i]=-2;
		for(int i=1;i<=m;i++) {
			int x,y; cin>>x>>y;
			if (x==y) b[x]=true;
			g1[x].push_back(y);
			g2[y].push_back(x);
		}
		for(int i=1;i<=n;i++) if (!vis[i]) dfs1(i);
		while (!s.empty()) {
			int t=s.top(); s.pop();
			if (vis[t]) {
				++len; dfs2(t);
			}
		}
		for(int i=1;i<=n;i++) {
			if (b[i]) {
				c[color[i]]=true;
			}
		}
		for(int i=1;i<=n;i++)
			for(int j=0;j<g1[i].size();j++) {
				int v=g1[i][j];
				if (color[i]!=color[v]) {
					g3[color[i]].push_back(color[v]);
					ru[color[v]]++;
				}
			}
		for(int i=1;i<=len;i++) {
			if (!ru[i]&&i!=color[1]) {
				f[i]=0;
				for(int j=0;j<g3[i].size();j++) {
					int v=g3[i][j];
					ru[v]--;
				}
			}
		}
		f[color[1]]=1;
		for(int i=1;i<=len;i++) if ((siz[i]>1||c[i])&&f[i]!=0) f[i]=-1;
		q.push(color[1]);
		while (!q.empty()) {
			int t=q.front(); q.pop(); 
			for(int i=0;i<g3[t].size();i++) {
				int v=g3[t][i];
				ru[v]--;
				if (f[v]!=-1) {
					if (f[t]==-1) f[v]=-1;
					else if (f[v]==-2&&f[t]==1) f[v]=1;
					else f[v]=2;
				}
				if (!ru[v]) {
					q.push(v);
				}
			}
		}
		for(int i=1;i<=n;i++) {
			cout<<f[color[i]]<<" ";
		}
		cout<<endl;
	}
	return 0;
}