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 <iostream>
#include <cstring>
#include <queue>
#include <vector>

const int NMAX = 4e5;

std::vector<int> graph[1 + NMAX];
std::vector<int> reverse[1 + NMAX];

int deg[1 + NMAX];

bool vis[1 + NMAX];
bool active[1 + NMAX];

int ans[1 + NMAX];

std::queue<int> q;

void dfsLoops(int node) {
  vis[node]    = true;
  active[node] = true;

  for (auto& nb: graph[node]) {
    ++deg[nb];

    if (!vis[nb])
      dfsLoops(nb);
    else if (active[nb]) {
      ans[nb] = -1;
      --deg[nb];
    }
  }

  active[node] = false;
}

/*
void dag(int root) {
  if (ans[root] != -1)
    ans[root] = 1;
  deg[1] = -1;

  q.push(root);

  while (!q.empty()) {
    auto node = q.front();
    q.pop();

    if (ans[node] != -1) {
      for (auto& i: reverse[node]) {
        if (ans[i] == -1) {
          ans[node] = -1;
          break;
        }

        if (ans[i] == 2)
          ans[i] = 2;
        else if (ans[i] == 1 && ans[node] != 2)
          ++ans[node];
      }
    }

    for (auto& nb: graph[node]) {
      --deg[nb];

      if (deg[nb] == 0)
        q.push(nb);
    }
  }
}
*/

void dfsSol(int node) {
  vis[node] = true;

  for (auto& nb: graph[node]) {
    int old = ans[nb];

    if (ans[nb] != -1) {
      if (ans[node] == -1)
        ans[nb] = -1;
      else if (ans[node] == 2)
        ans[nb] = 2;
      else if (ans[node] == 1 && ans[nb] < 2)
        ++ans[nb];
    }

    if (!vis[nb] || old != ans[nb])
      dfsSol(nb);
  }
}

int main() {
  int tests;
  std::cin >> tests;

  int n, m;
  while (tests--) {
    std::cin >> n >> m;

    for (int i = 1; i <= n; ++i) {
      graph[i].clear();
      reverse[i].clear();
      deg[i]    = 0;
      ans[i]    = 0;
      vis[i]    = false;
      active[i] = false;
    }

    for (int i = 1; i <= m; ++i) {
      int x, y;
      std::cin >> x >> y;

      graph[x].push_back(y);
      reverse[y].push_back(x);
    }

    dfsLoops(1);

    memset(vis, 0, sizeof(vis));

    if (ans[1] != -1)
      ans[1] = 1;

    dfsSol(1);

    //dag(1);

    for (int i = 1; i <= n; ++i)
      std::cout << ans[i] << ' ';
    std::cout << '\n';
  }

  return 0;
}