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 fio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define en cin.close();return 0;
#define INF ((ll)(2e9))
#define pb push_back
#define fi first//printf("%lli\n",cur);
#define se second//scanf("%lli",&n);
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 4e5+5;
vector<int> g[N];
int col[N];
bool mark[N][2];
void dfs(int u)
{
    col[u]=1;
    for(auto v:g[u])
    {
        if(!col[v])
            dfs(v);
        else if(col[v]==1)
            mark[v][0]=1;
        else
            mark[v][1]=1;
    }
    col[u]=2;
}
void dfs2(int u, int d)
{
    mark[u][d]=1;
    for(auto v:g[u])
        if(!mark[v][d])
            dfs2(v,d);
}
int main()
{
    fio
    //ifstream cin("in.inn");
    int qq;
    cin >> qq;
    while(qq--)
    {
        int n, m;
        cin >> n >> m;
        while(m--)
        {
            int u, v;
            cin >> u >> v;
            g[u].pb(v);
        }
        dfs(1);
        for(int i = 1;i<=n;i++)
        {
            if(mark[i][0])
                dfs2(i,0);
            if(mark[i][1])
                dfs2(i,1);
        }
        for(int i = 1;i<=n;i++)
        {
            if(!col[i])
                cout << 0;
            else if(mark[i][0])
                cout << -1;
            else if(mark[i][1])
                cout << 2;
            else
                cout << 1;
            cout << " ";
        }
        cout << "\n";
        for(int i = 1;i<=n;i++)
            g[i].clear(),
            col[i]=0,
            mark[i][0]=0,
            mark[i][1]=0;
    }
    //cin.close();
    return 0;
}