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>

using namespace std;

#define ull unsigned long long
#define ll long long
#define ui unsigned int
#define us unsigned short
#define inf_int 1e9
#define inf_ll 1e18
#define mod 1000000007
#define smod 998244353

const int maxN = 4e5 + 5;

vector<pair<int, int>> mark[3]; // 0 multiple path, 1 self loop, 2 cycle
vector<int> adj[maxN];
int ans[maxN];

void dfs(int curNode, bool visited[], bool cur[]){
	visited[curNode] = true;
	cur[curNode] = true;
		
	for(auto nxt : adj[curNode]){
		if(nxt != curNode){
			if(visited[nxt]){
				if(cur[nxt]){
					mark[2].push_back({nxt, 2});
				} else {
					mark[0].push_back({nxt, 1});
				}
			} else {
				dfs(nxt, visited, cur);
			}
		} else {
			mark[1].push_back({nxt, 0});
		}
	}
	
	cur[curNode] = false;
}

void dfs2(int id, int curNode, bool visited[]){
	if(id == -1){
		ans[curNode] = 1;
	} else if(id == 0){
		ans[curNode] = 2;
	} else if(id == 2){
		ans[curNode] = -1;
	} else if(id == 1){
		ans[curNode] = -1;
	}
	
//	cout << id << " " << curNode << "\n";
	visited[curNode] = true;
	for(auto nxt : adj[curNode]){
		if(!visited[nxt]){
			dfs2(id, nxt, visited);
		}
	}
}

int main(){
	ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
	
	us t;
	cin >> t;
	
	while(t--){
		int n, m;
		cin >> n >> m;
		
		for(int x=0;x<n;x++) adj[x].clear();
		for(int x=0;x<3;x++) mark[x].clear();
		for(int x=0;x<m;x++){
			int a, b;
			cin >> a >> b;
			
			a--; b--;
			adj[a].push_back(b);
		}
		
		bool visited[n], cur[n];
		for(int x=0;x<n;x++){
			visited[x] = cur[x] = false;
		}
		
		dfs(0, visited, cur);
		
		for(int x=0;x<n;x++){
			ans[x] = 0;
		}
		
		fill(visited, visited+n, false);
		dfs2(-1, 0, visited);
		
		for(int id=0;id<3;id++){
			fill(visited, visited+n, false);
			
			for(auto p : mark[id]){
//				cout << p.first << "\n";
				dfs2(id, p.first, visited);
			}
		}
		
		for(int x=0;x<n;x++){
			if(x > 0) cout << " ";
			cout << ans[x];
		}
		cout << "\n";
	}

    return 0;
}