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<algorithm>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
const int maxn=4e5;
int T,n,m,x,y,tag,newn,dfn[maxn+1],low[maxn+1],stk[maxn+1],belong[maxn+1],size[maxn+1],degree[maxn+1];
bool self[maxn+1],inf[maxn+1],multi[maxn+1],in_stack[maxn+1];
vector<int> edge[2][maxn+1]; 
queue<int> que;
inline void pop_stack(void)
{
	in_stack[stk[stk[0]]]=false;
	belong[stk[stk[0]]]=newn;
	++size[newn];
	if(self[stk[stk[0]]])
		++size[newn];
	--stk[0];
	return;
}
void Tarjan(int now)
{
	stk[++stk[0]]=now;
	low[now]=dfn[now]=++tag;
	in_stack[now]=true;
	for(int i=0;i<edge[0][now].size();++i)
		if(!dfn[edge[0][now][i]])
		{
			Tarjan(edge[0][now][i]);
			low[now]=min(low[now],low[edge[0][now][i]]);
		}
		else if(in_stack[edge[0][now][i]])
			low[now]=min(low[now],dfn[edge[0][now][i]]);
	if(dfn[now]==low[now])
	{
		size[++newn]=0;
		while(stk[stk[0]]!=now)
			pop_stack();
		pop_stack();
	}
	return;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		newn=tag=0;
		for(int i=1;i<=n;++i)
		{
			multi[i]=degree[i]=inf[i]=dfn[i]=self[i]=0;
			edge[0][i].clear(); 
			edge[1][i].clear();
		}		
		for(int i=1;i<=m;++i)
		{
			scanf("%d%d",&x,&y);
			if(x!=y)
				edge[0][x].push_back(y);
			else self[x]=true;
		}
		Tarjan(1);
		for(int i=1;i<=n;++i)
			if(dfn[i])
				for(int j=0;j<edge[0][i].size();++j)
					if(dfn[edge[0][i][j]]&&belong[i]&&belong[edge[0][i][j]]&&belong[i]!=belong[edge[0][i][j]])
					{
						++degree[belong[edge[0][i][j]]];
						edge[1][belong[i]].push_back(belong[edge[0][i][j]]);
					}	
		for(int i=1;i<=n;++i)
			if(degree[i]>1)
				multi[i]=true;	
		while(!que.empty())
			que.pop();
		if(size[belong[1]]>1)
			inf[belong[1]]=true;
		que.push(belong[1]);
		while(!que.empty())
		{
			int front=que.front();
			que.pop();
			if(size[front]>1)
				inf[front]=true;
			for(int i=0;i<edge[1][front].size();++i)
			{
				--degree[edge[1][front][i]];
				multi[edge[1][front][i]]|=multi[front];
				inf[edge[1][front][i]]|=inf[front];
				if(!degree[edge[1][front][i]])
					que.push(edge[1][front][i]);
			}
		}
		for(int i=1;i<=n;++i)
			if(!dfn[i])
				printf("0 ");
			else if(inf[belong[i]])
				printf("-1 ");
			else if(multi[belong[i]])
				printf("2 ");
			else printf("1 ");
		puts("");
	}
	return 0;
}