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:
/*
K.D. Vinit  /,,/
*/
#include<bits/stdc++.h>
using namespace std;

#define int long long
const int N = 4e5 + 1;

vector<int> adj[N];
int vis1[N], vis2[N], vis3[N], cur[N], ans[N], Infi[N];

void DFS(int chi)
{
    ans[chi]++; vis1[chi]=1; cur[chi]=1;
    for(auto x: adj[chi])
    {
        if(vis1[x]==1) { ans[x]++; if(cur[x]==1) Infi[x]=1; }
        else DFS(x);
    }
    cur[chi]=0;
}

void DFS2(int chi)
{
    vis2[chi]=1; Infi[chi]=1;     ans[chi]=-1;
    for(auto x: adj[chi]) if(!vis2[x]) DFS2(x);
}

void DFS3(int chi)
{
    vis3[chi]=1; ans[chi]=2;
    for(auto x: adj[chi]) if(!vis3[x]) DFS3(x);
}

void solve()
{
    int n, m; cin>>n>>m;
    for(int i=1; i<=n; i++) { adj[i].clear(); vis1[i]=0; vis2[i]=0; cur[i]=0; ans[i]=0; Infi[i]=0; vis3[i]=0; }
    for(int i=1; i<=m; i++) { int u, v; cin>>u>>v; adj[u].push_back(v); }
    DFS(1);
    for(int i=1; i<=n; i++) if(ans[i]>=2 && vis3[i]==0) DFS3(i);
    for(int i=1; i<=n; i++) if(Infi[i]==1 && vis2[i]==0) DFS2(i);
    for(int i=1; i<=n; i++) cout<<ans[i]<<" "; cout<<endl;
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t; cin>>t;
    while(t--) solve();
    return 0;
}