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;

typedef long long ll;
typedef pair<ll,ll> pi;
typedef vector <ll> vi;
typedef vector <pi> vpi;
typedef pair<pi,ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0)
#define maxn 400010

int N,M,a,b;
vi adj[maxn];
int vis[maxn], ans[maxn];

void dfs(int x){
	if (vis[x] == 2){
		if (ans[x] != -1) ans[x] = 2;
		return;
	}
	if (vis[x] == 1){
		ans[x] = -1; return;
	}
	vis[x] = 1;
	
	aFOR(i,adj[x]){
		dfs(i);
	}
	vis[x] = 2;
}
void dfs2(int x){
	ans[x] = -1;
	aFOR(i,adj[x]) if (ans[i] != -1) dfs2(i);
}
void dfs3(int x){
	ans[x] = 2;
	aFOR(i,adj[x]) if (ans[i] != 2 && ans[i] != -1) dfs3(i);
}
void solve(){
	cin >> N >> M;
	FOR(i,1,N) adj[i].clear();
	mem(ans, 0);
	mem(vis, 0);
	FOR(i,1,M){
		cin >> a >> b; adj[a].pb(b);
	} 
	
	dfs(1);
	FOR(i,1,N) if (ans[i] == -1) dfs2(i);
	FOR(i,1,N) if (ans[i] == 2) dfs3(i);
	FOR(i,1,N){
		if (ans[i] == 0){
			if (vis[i] == 0) cout << 0 << ' ';
			else cout << 1 << ' ';
		}else cout << ans[i] << ' ';
	}
	cout << '\n';
}
int main(){
	fast;
	int TC;
	cin >> TC;
	while (TC--) solve();
	
}