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;

vector<vector<int>> g, gt, scc;
vector<int> order, vis, com;

void dfs1 (int v) {
  vis[v]=1;
  for (int u : g[v]){
    if (!vis[u])  dfs1 (u);
  }
  order.push_back (v);
}

void dfs2(int v){
  vis[v]=1;
  com.push_back(v);
  for(int u: gt[v]){
    if(!vis[u]) dfs2(u);
  }
}


int main(){
  ios::sync_with_stdio(0), cin.tie(0);
  
  int T; cin >> T;
  while(T--){
    int n, m;  cin >> n >> m;
    g.clear(), gt.clear();
    g.resize(n), gt.resize(n);
    vector<int> slef_loop(n);
    for (int i = 0; i < m; ++i){
      int u, v;  cin >> u >> v;  u--, v--;
      if (u==v) {
        slef_loop[u] = 1;
      } else {
        g[u].push_back(v);
        gt[v].push_back(u);
      }
    }


    vis.assign(n, 0);
    order.clear();
    for (int i = 0; i < n; ++i){
      if(!vis[i])
        dfs1(i);
    }
    vis.assign(n, 0);
    scc.clear();
    for (int i = 0; i < n; ++i){
      int v=order[n-1-i];
      if(!vis[v]){
        dfs2(v);
        scc.push_back(com);
        com.clear();
      }
    }


    int sz = scc.size();
    int comp_no[n];
    vector<int> slef_loop_comp(sz, 0);
    for (int i = 0; i < sz; ++i){
      for (int& u: scc[i]) {
        comp_no[u] = i;
        if (slef_loop[u]) {
          slef_loop_comp[i] = 1;
        }
      }
    }

    set<int> scc_g[sz];
    for (int u = 0; u < n; ++u){
      for (int& v: g[u]) {
        if (comp_no[u] != comp_no[v]) scc_g[comp_no[u]].insert(comp_no[v]);
      }
    }
    
    queue<int> q;  q.push(comp_no[0]);
    vector<int> tp(sz, 0);  tp[comp_no[0]] = 1;

    while(!q.empty()) {
      int u = q.front();  q.pop();
      if (scc[u].size()>1 or slef_loop_comp[u])  tp[u] = -1;
      for (int v: scc_g[u]) {
        if (tp[u]==-1 and tp[v]!=-1) {
          tp[v] = -1;
          q.push(v);
        } else if (tp[u]==2 and tp[v]!=2 and tp[v]!=-1) {
          tp[v] = 2;
          q.push(v);
        } else if (tp[u]==1 and tp[v]==1) {
          tp[v] = 2;
          q.push(v);
        } else if (tp[u]==1 and tp[v]==0) {
          tp[v] = 1;
          q.push(v);
        }
      }
    }
    
    for (int i = 0; i < n; ++i){
      cout << tp[comp_no[i]] << " ";
    }
    cout << '\n';
  }
  
  return 0;
}