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<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<cmath>
#include<set>
#include<iomanip>
#include<map>
#include<stdio.h>
#define INF 0x3f3f3f3f
#define ll long long
#define pll pair<ll,ll>
#define pdl pair<double,ll>
#define pdd pair<double, double>
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 
using namespace std;
const int N = 4e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-4;
ll n,m;
vector<ll>g[N];
int color[N];
bool circle[N];
bool mul[N];
int ans[N];
bool vis[N];
void dfs_color(ll cur)
{
	color[cur]=1;
	for(auto i:g[cur])
	{
		if(color[i]==1)//若当前节点被第二次访问时其颜色是1，说明它是一个带环节点
		{
			circle[i]=1;
		}
		else if(color[i]==2)//若当前节点被第二次访问时其颜色是2，说明它是一个多路径节点
		{
			mul[i]=1;
		}
		else if(color[i]==0)
		{
			dfs_color(i);
		}
	}
	color[cur]=2;
}
void dfs_mul(ll cur,int m)
{
	vis[cur]=1;
	if (mul[cur] == 1)
	{
		m = 1;
	}
	if(m==1)
	{
		ans[cur]=2;
	}
	
	//cout<<"cur is "<<cur<<" cir=="<<cir<<" m=="<<m<<" ans=="<<ans[cur]<<" color=="<<color[cur]<<endl;
	for(auto i:g[cur])
	{
		if(!vis[i]){
			dfs_mul(i,m);
		}

	}
}
void dfs_cir(ll cur,ll cir)
{
	vis[cur]=1;
	if (circle[cur] == 1)
	{
		cir = 1;
	}
	if(cir==1)
	{
		ans[cur]=-1;
	}
	
	for(auto i:g[cur])
	{
		if(!vis[i]){
			dfs_cir(i,cir);
		}

	}
}
void dfs_reachable(ll cur)
{
	vis[cur]=1;
	ans[cur]=1;
	for(auto i:g[cur])
	{
		if(!vis[i]){
			dfs_reachable(i);
		}

	}
}
void solve()
{
	cin>>n>>m;
	for(ll i=1;i<=n;i++)
	{
		g[i].clear();
		color[i]=0;//每个节点的颜色，0为未访问，1为访问过但没遍历其所有后继节点，2为完全访问了其所有后继节点
		circle[i]=0;//该节点是不是环路节点
		mul[i]=0;//该节点是不是多路径节点
		ans[i]=0;
		vis[i]=0;
	}
	for(ll i=0;i<m;i++)
	{
		ll a,b;
		cin>>a>>b;
		g[a].push_back(b);
	}
	dfs_color(1);
	memset(vis,0,sizeof vis);
	dfs_reachable(1);//检验每个节点的可达性
	/*为什么不把可达性、多路径、环放在一个dfs里检验呢？
	这是因为放在一个里面可能导致后面的结果覆盖前面的结果，而我们发现环的优先级大于多路径优先级，多路径优先级又大于单路径优先级
	所以分为三个dfs，优先级大的就可以覆盖优先级小的，而不会造成混乱
	*/
	memset(vis,0,sizeof vis);
	for (ll i = 1; i <= n; i++)
	{
		if (mul[i] == 1)
		{
			dfs_mul(i, mul[i]);//找出所有多路径节点的可达节点
			/*
			注意这里要遍历每个多路径节点（下面判环时也要遍历每个带环节点），如果只用一次dfs去标记的话，可能会出现一个多路径节点标号比较靠后，从而它后面的节点没有标记到的情况
			比如说下面这组数据：
				5 6
				1 2
				2 3
				1 4
				4 3
				4 4
				3 5
			如果只进行一次dfs就会出现错误（这组数据是在判环时会出错）
			*/
		}
	}
	memset(vis,0,sizeof vis);
	for (ll i = n; i >=1; i--)
	{
		if (circle[i] == 1)
		{
			dfs_cir(i,circle[i]);
		}
	}
	
	for(ll i=1;i<=n;i++)
	{
		cout<<ans[i]<<' ';
	}
	cout<<endl;
	
}
int main()
{
	IOS;
	ll _;
	cin>>_;
	while(_--)
	solve();

	
}