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>
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define pii pair<int, int>
#define pll pair<long long,long long>
#define vl vector<long long>
#define vll vector<pll>
#define vi vector<int>
#define vii vector<pii>
#define sws ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl '\n'
#define tcase int tt; cin>>tt; while(tt--)
#define tcase2 int tt; cin>>tt; for(int qq=1;qq<=tt;qq++)
using namespace std;
typedef long long int ll;
typedef  unsigned long long int ull;
typedef long double ld;
const int MAXN = 1000002;
const int MOD2 = 998244353;
const int MOD = 1000000007; 
const int INF = 1e8;
const ld EPS = 1e-7;
// Extra
#define forn(i, n) for(int i = 0; i < (int)n; i++)
#define forne(i, a, b) for(int i = a; i <= b; i++)
#define all(x) x.begin(), x.end()
#define trav(a, x) for(auto& a : x)
#define fill(x,y) memset(x,y,sizeof(x))
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

ll mul(ll x, ll y)
{
    return (x * 1ll * y) % MOD;
}
 
ll fastpow(ll x, ll y)
{
    ll z = 1;
    while(y)
    {
        if(y & 1) z = mul(z, x);
        x = mul(x, x);
        y >>= 1;
    }
    return z;
}
ll modinv(ll n,ll p)
{
	return fastpow(n,p-2) ;
}
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first < b.first;
    }
    return a.second >b.second;
  }
};
ll ans[400001] ;
vector<ll>adj[400001] ;
int vis[400001] ;
queue<ll>qa,qb ;
void dfs(ll x)
{
	if(vis[x]==1)
	{   ans[x]=-1 ;
		qa.push(x) ;
		return  ;
	}
	if(vis[x]==2)
	{   if(ans[x]!=-1)
	    ans[x]=2 ;
		qb.push(x) ;
		return ;
	}
//	cout<<x<<" ";
	vis[x]=1 ;
	if(ans[x]!=-1)
	ans[x]=max(ans[x],(ll)1) ;
	for(auto u:adj[x])
	dfs(u) ;
	vis[x]=2 ;
}
int main(int argc, char** argv)
{   sws ;
//  cout<<setprecision(15);
 ll google=1 ; 
   tcase
    {  
    ll n,m ;
    cin>>n>>m ;
    forn(i,n+1)
    {
    	vis[i]=0 ;
    	adj[i].clear() ;
    	ans[i]=0 ;
	}
    for(int i=0;i<m;i++)
    {
    	ll u,v ;
    	cin>>u>>v; 
    	adj[u].pb(v) ;
	}
	dfs(1)  ;
	forn(i,n+1)
	vis[i]=0 ;
	while(!qa.empty())
	{
		ll x=qa.front() ;
		qa.pop() ;
		if(vis[x])
		continue ;
		vis[x]=1 ;
		for(auto u:adj[x])
		{   if(ans[u]!=-1)
		   { ans[u]=-1 ;
			qa.push(u);
		    }
		}
	}
	forn(i,n+1)
	vis[i]=0 ;
	while(!qb.empty())
	{
		ll x=qb.front() ;
		qb.pop() ;
		if(vis[x])
		continue ;
		if(ans[x]==-1)
		continue ;
		vis[x]=1 ;
		for(auto u:adj[x])
		{  if(ans[u]==-1)
		   continue ; 
		   if(ans[u]!=2)
		   { ans[u]=2;
			qb.push(u);
		    }
		}
	}
	for(int i=1;i<=n;i++)
	cout<<ans[i]<<" ";
	cout<<"\n" ;
    }
    
return 0 ;
}