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;

struct nodS
{
    int viz;
    bool dublu, inf;
    vector<int> v;
};

int n,m,t,i;
nodS nod[400005];

void dfs(int p)
{
    nod[p].viz=1;
    for(int it : nod[p].v)
    {
        if(!nod[it].viz)
            dfs(it);
        else if(nod[it].viz==1)
            nod[it].inf=true;
        else
            nod[it].dublu=true;
    }
    nod[p].viz=2;
}
void pushInf(int p)
{
    nod[p].inf=true;

    for(int it : nod[p].v)
        if(!nod[it].inf)
            pushInf(it);
}
void pushDublu(int p)
{
    nod[p].dublu=true;

    for(int it : nod[p].v)
        if(!nod[it].dublu)
            pushDublu(it);
}

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

    cin>>t;
    while(t)
    {
        t--;
        cin>>n>>m;

        for(i=1;i<=n;i++)
        {
            nod[i].v.clear();
            nod[i].viz=0;
            nod[i].dublu=false;
            nod[i].inf=false;
        }

        for(i=1;i<=m;i++)
        {
            int x, y;
            cin>>x>>y;
            nod[x].v.push_back(y);
        }

        dfs(1);

        for(i=1;i<=n;i++)
        {
            if(!nod[i].viz)
                continue;

            if(nod[i].inf)
                pushInf(i);
            if(nod[i].dublu)
                pushDublu(i);
        }

        for(i=1;i<=n;i++)
        {
            if(!nod[i].viz)
                cout<<0;
            else if(nod[i].inf)
                cout<<-1;
            else if(nod[i].dublu)
                cout<<2;
            else
                cout<<1;

            cout<<' ';
        }
        cout<<'\n';
    }
    return 0;
}