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;

#define int long long
#define fi first
#define se second
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define mp make_pair
//#define endl "\n"
#define ll long long
#define ld long double
//#define pair<int,int> p32
//#define pair<ll,ll> p64
//#define pair<double,double> pdd
//#define vector<int> v32
//#define vector<ll> v64
//#define vector<string> vs
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
ll mod=998244353;
vector<int> ans,used;
vector<bool> p;
vector<vector<int> > g;
int n,m;

void dfs(int v){
    used[v]=1;

    for(auto to:g[v]){
        if(used[to]==1){
            p[to]=true;
        }else if(used[to]==0){
            dfs(to);
        }
    }

    used[v]=2;
}

void dfs2(int v){
    used[v]=1;
    ans[v]++;

    for(auto to:g[v]){
        if(p[to]){
            continue;
        }else if(ans[to]<=2){
            dfs2(to);
        }
    }

    used[v]=2;
}

void dfs3(int v){
    used[v]=1;
    ans[v]=-1;

    for(auto to:g[v]){
        if(p[to]==1){
            continue;
        }else if(used[to]==0){
            dfs3(to);
        }
    }

    used[v]=2;
}


signed main(){
    fast_cin();

    int t;
    cin>>t;

    forn(_i,t){
        cin>>n>>m;

        g.assign(n,vector<int> ());
        for(int i=0;i<m;i++){
            int a,b;
            cin>>a>>b;
            a--;b--;

            g[a].push_back(b);
//            g[b].push_back(a);
        }

        p.assign(n,false);
        ans.assign(n,0);
        used.assign(n,0);
        dfs(0);

        used.assign(n,0);
        dfs2(0);

        used.assign(n,0);
        auto p2=p;
        for(int i=0;i<n;i++){
            if(p[i]){
                dfs3(i);
            }
        }
        auto ans2=ans;

        for(auto i:ans){
            cout<<min(i,2ll)<<" ";
        }
        cout<<endl;
    }

}