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 monke_flip ios_base::sync_with_stdio(false); cin.tie(NULL);
#define random_monke srand(chrono::system_clock::now().time_since_epoch().count());
#ifdef LEL
#include <dbg.h>
#else
#define dbg(...) {/* temon kichu na; */}
#endif

using namespace std;
using ll = long long;
const int MONKE = 0;

class SCC
{
    private:
    vector <bool> vis;
    vector <int> order, component,idx;
    vector <vector <int> > adj, rev,COMPONENTS,DAG;
    vector <set<int> > _DAG;
    void init(int n,vector <pair<int,int> > &edges){
        adj.clear(),rev.clear(), COMPONENTS.clear(),DAG.clear(),_DAG.clear();
        adj.resize(n),rev.resize(n);
        for(auto &e:edges){
            adj[e.first].push_back(e.second);
            rev[e.second].push_back(e.first);
        }
    }
    void dfs1(int node){
        vis[node] = true;
        for(auto x:adj[node]){
            if(!vis[x]) dfs1(x);
        }
        order.push_back(node);
    }

    void dfs2(int node){
        vis[node] = true;
        component.push_back(node);
        for(auto x:rev[node]){
            if(!vis[x]) dfs2(x);
        }
    }
    public:
    vector <vector <int> > get_SCC(int n,vector <pair<int,int> > &edges){
        init(n,edges);
        vis.assign(n,false);
        order.clear();
        for(int i=0;i<n;i++){
            if(!vis[i]){
                dfs1(i);
            }
        }
        vis.assign(n,false);
        idx.assign(n,0);
        reverse(order.begin(), order.end());
        int i = 0;
        for(auto v:order){
            component.clear();
            if(!vis[v]) dfs2(v);
            if(!component.empty()){
                COMPONENTS.push_back(component);
                for(auto x:component){
                    idx[x] = i;
                }
                i++;
            }
        }
        int sz = COMPONENTS.size();
        _DAG.resize(sz), DAG.resize(sz);
        for(auto x:edges){
            _DAG[idx[x.first]].insert(idx[x.second]);
        }
        for(int i=0;i<sz;i++){
            for(auto x:_DAG[i]){
                if(x==i) continue;
                DAG[i].push_back(x);
            }
        }
        return COMPONENTS;
    }
    vector <int> get_idx(){
        return idx;
    }
    vector <vector <int> > get_DAG(){
        return DAG;
    }
};

void solve(){
    int _n, m;
    cin >> _n >> m;
    SCC G;
    set <int> self;
    vector <pair<int,int>> edges(m);
    for(auto &[u,v]:edges){
        cin >> u >> v;
        u--, v--;
        if(u == v)
            self.insert(u);
    }
    vector <vector <int> > scc, DAG;
    vector <int> new_idx;
    
    scc = G.get_SCC(_n,edges);
    DAG = G.get_DAG();
    new_idx = G.get_idx();

    int n = scc.size();
    vector <bool> inf(n);
    for(int i=0; i<n; i++){
        inf[i] = (int(scc[i].size()) > 1);
    }
    for(auto x:self){
        inf[new_idx[x]] = true;
    }
    vector <vector <int> > REV(n);
    for(int u = 0; u<n; u++){
        for(auto v:DAG[u]){
            REV[v].push_back(u);
        }
    }

    int root = new_idx[0];
    vector <int> dp(n);
    vector <bool> vis(n);
    dp[root] = (inf[root] ? -1:1);
    vis[root] = true;

    function <int(int)> call = [&](int node){
        if(vis[node])
            return dp[node];
        vis[node] = true;
        int cnt1 = 0, cnt2 = 0;
        for(auto x:REV[node]){
            int c = call(x);
            c == -1? cnt2++: cnt1 += c;
        }
        if(cnt2 == 0 and cnt1 == 0){
            return dp[node] = 0;
        }
        if(inf[node] or cnt2 > 0)
            return dp[node] = -1;
        return dp[node] = (cnt1 > 1 ? 2:1);
    };
    vector <int> ans(_n);
    for(int i=0;i<n;i++){
        int val = call(i);
        for(auto x:scc[i])
            ans[x] = val;
    }
    for(auto x:ans)
        cout<<x<<' ';
    cout<<'\n';
}

int main()
{
    monke_flip
    int t = 1; 
    cin>>t;
    for(int tc=1; tc<=t; tc++){
        solve();
    }
    return MONKE;
}