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;
long long mod = 998244353;
long long n , m , dp [400005] , wt [400005] , num [400005] , vis [400005];
vector < long long > gr [400005] , ngr [400005];
void dfs ( int x )
{
    if ( wt [x] ) return;
    vis [x] = 1;
    for ( int i = 0 ; i < ngr [x] . size () ; i ++ )
    {
        int u = ngr [x][i];
        if ( vis [u] ) continue;
        dp [u] += dp [x];
        if ( dp [u] < 1e9 ) dp [u] = min ( dp [u] , 4ll );
        num [u] --;
        if ( num [u] < 0 ) cout << 0 / 0;
        if ( num [u] == 0 ) dfs ( u );
    }
}
void go ( int x )
{
    vis [x] = 1;
    wt [x] = 1;
    for ( int i = 0 ; i < ngr [x] . size () ; i ++ )
    {
        int u = ngr [x][i];
        if ( vis [u] ) continue;
        go ( u );
    }
}
void cf ( int x )
{
    vis [x] = 1;
    for ( int i = 0 ; i < gr [x] . size () ; i ++ )
    {
        int u = gr [x][i];
        ngr [x] . push_back ( u );
        num [u] ++;
        if ( vis [u] == 1 ) wt [x] = 1;
        else if ( vis [u] == 0 ) cf ( u );
    }
    vis [x] = 2;
}
int main()
{
    int t;
    cin >> t;
    while ( t -- )
    {
        cin >> n >> m;
        for ( int i = 0 ; i <= n ; i ++ )
        {
            gr [i] . clear () , ngr [i] . clear ();
            dp [i] = 0 , wt [i] = 0 , num [i] = 0 , vis [i] = 0;
        }
        for ( int i = 0 ; i < m ; i ++ )
        {
            int x , y;
            cin >> x >> y;
            gr [x] . push_back ( y );
        }
        cf ( 1 );
        for ( int i = 1 ; i <= n ; i ++ ) vis [i] = 0;
        for ( int i = 1 ; i <= n ; i ++ )
        {
            if ( vis [i] == 0 && wt [i] ) go ( i );
        }
        for ( int i = 1 ; i <= n ; i ++ ) vis [i] = 0;
        dp [1] = 1;
        dfs ( 1 );
        for ( int i = 1 ; i <= n ; i ++ )
        {
            if ( wt [i] ) cout << -1 << ' ';
            else if ( dp [i] > 1 ) cout << 2 << ' ';
            else cout << dp [i] << ' ';
        }
        cout << endl;
    }
}
/*
3
4 5
1 2
2 1
2 3
3 4
4 2



5 5
1 5
4 5
2 3
3 4
4 2


6 7
2 3
3 4
4 2
1 5
1 6
5 6
4 5
*/