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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <numeric>
#include <cmath>
using namespace std;
typedef long long ll;

#define finc(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define fdec(i,a,b) for(int i=(int)(b);i-->(int)(a);)
#define endb ' '

const int INF = 0xffff;
int n, m;
vector<int> ans;
vector<int> note;
queue<int> dot;
struct Graph
{
	int n = 0;
	int m = 0;
	vector<int> head;
	struct Edge
	{
		int next;
		int uDot;
		int vDot;
	};
	vector<Edge> edge;
	void init(int nn, int mm)
	{
		n = nn;
		m = 0;
		head = vector<int>(nn + 1);
		edge = vector<Edge>(mm + 1);
	}
	void addEdge(int u, int v)
	{
		m++;
		edge[m].next = head[u];
		edge[m].uDot = u, edge[m].vDot = v;
		head[u] = m;
	}
};
Graph gra, regra;

void check(int r)
{
	note[r] = 1;
	for (int i = gra.head[r]; i; i = gra.edge[i].next)
	{
		int v = gra.edge[i].vDot;
		if (note[v] == 0)
			check(v);
		else
			if (note[v] == 1)
				dot.push(v);
	}
	note[r] = 2;
}
void setInf()
{

	while (!dot.empty())
	{
		int I = dot.front();
		dot.pop();
		if (ans[I] == INF)
			continue;
		ans[I] = INF;
		for (int i = gra.head[I]; i; i = gra.edge[i].next)
			dot.push(gra.edge[i].vDot);
	}
}
int setAns(int r)
{
	if (ans[r] != -1)
		return ans[r];
	ans[r] = 0;
	for (int i = regra.head[r]; i; i = regra.edge[i].next)
	{
		int v = regra.edge[i].vDot;
		if (setAns(v) != INF)
		{
			ans[r] = min(2, ans[r] + ans[v]);
			if (ans[r] > 1)
				break;
		}
		else
		{
			ans[r] = INF;
			break;
		}
	}
	return ans[r];
}

void ac(void)
{
	cin >> n >> m;
	ans = vector<int>(n + 1, -1), note = vector<int>(n + 1), dot = queue<int>();
	gra.init(n, m), regra.init(n, m);
	finc(i, 1, m + 1)
	{
		int u, v; cin >> u >> v;
		gra.addEdge(u, v), regra.addEdge(v, u);
	}
	check(1);
	setInf();
	if (ans[1] != INF)
		ans[1] = 1;
	finc(i, 1, n + 1)
		setAns(i);
	finc(i, 1, n + 1)
		cout << ((ans[i] == INF) ? (-1) : (ans[i])) << endb;
	cout << endl;
}
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int test = 1; cin >> test;
	while (test--)
		ac();

	return 0;
}