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;
const int nax = 4e5 + 5;
queue<int> cue;
vector<int> edge[nax];
vector<int> is(nax), vis(nax), in(nax);
void find_cycle (int cur) {
  vis[cur] = 1;
  if (is[cur] == 2) return ;
  if (is[cur] == 1) {
    cue.push(cur);
    in[cur] = 1;
    return ;
  }
  is[cur] = 1;
  for (int to : edge[cur]) {
    find_cycle(to);
  }
  is[cur] = 2;
}
void process_cue (int value) {
  while (not cue.empty()) {
    int elm = cue.front();
    cue.pop();
    vis[elm] = value;
    for (int to : edge[elm]) {
      if (in[to] == 0 && vis[to] == 1) {
        cue.push(to);
        in[to] = 1;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int test;
  cin >> test;
  while (test--) {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      edge[i].clear();
    }
    for (int i = 1; i <= m; i++) {
      int x, y;
      cin >> x >> y;
      edge[x].push_back(y);
    }
    fill(vis.begin() + 1, vis.begin() + 1 + n, 0);
    fill(is.begin() + 1, is.begin() + 1 + n, 0);
    fill(in.begin() + 1, in.begin() + 1 + n, 0);
    find_cycle(1);
    process_cue(-1);
    vector<int> cnt(n + 1, 0);
    fill(in.begin() + 1, in.begin() + 1 + n, 0);
    for (int i = 1; i <= n; i++) {
      if (vis[i] != 1) continue;
      for (int to : edge[i]) {
        if (vis[to] != 1) continue;
        if (++cnt[to] == 2) {
          cue.push(to);
          in[to] = 1;
        }
      }
    }
    process_cue(2);
    for (int i = 1; i <= n; i++) {
      cout << vis[i] << " \n" [i == n];
    }
  }
  return 0;
}