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>

typedef long long ll;
const int N = 4e5+5;
std::vector<int> g[N];
int ans[N];
bool visited[N];
bool visited2[N];
bool visited3[N];
bool cycle[N];
std::pair<int,int> timer[N];
int n,m;
int ct;	//curr time;

void dfs(int u, int p){
	timer[u].first = ct++;
	timer[u].second = INT_MAX;
	visited[u] = true;
	if(ans[u] == 0) {
		ans[u] = 1;
	}

	for(int v:g[u]) {
		if(visited[v]) {
			if(timer[v].second == INT_MAX) {
				ans[v] = -1;
				cycle[v] = true;
			} else if(!cycle[v]){
				ans[v] = 2;
			}
		} else {
			dfs(v, u);
		}
	}
	timer[u].second = ct++;
}

void dfs2(int u, int p){
	visited2[u] = true;
	ans[u] = -1;
	for(int v:g[u]) {
		if(!visited2[v]) {
			ans[v] = -1;
			cycle[v] = true;
			dfs2(v, u);
		}
	}
}

void dfs3(int u, int p){
	visited3[u] = true;
	if(ans[u] == 1){
		ans[u] = 2;
	}
	for(int v:g[u]) {
		if(!visited3[v]) {
			// ans[v] = 2;
			// cycle[v] = true;
			dfs3(v, u);
		}
	}
}


int main(){
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(0);

	int t; std::cin >> t;
	while(t--) {
		std::cin >> n >> m;

		ct = 0;
		for(int i=0;i<n;i++) {
			visited[i] = false;
			visited2[i] = false;
			visited3[i] = false;
			cycle[i] = false;
			ans[i] = 0;
			g[i].clear();
		}

		for(int i=0;i<m;i++) {
			int u,v;
			std::cin >> u >> v;
			u--;v--;
			g[u].push_back(v);
		}
		ans[0] = 1;
		dfs(0,0);
		for(int i=0;i<n;i++) {
			if(!visited2[i] && cycle[i]) {
				dfs2(i, i);
			}
		}
		for(int i=0;i<n;i++) {
			if(!visited3[i] && ans[i] == 2) {
				dfs3(i, i);
			}
		}

		for(int i=0;i<n;i++) {
			std::cout << ans[i] << " ";
		}
		std::cout << std::endl;
	}

	return 0;
}