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;


void dfs3(int v, vector<vector<int>> &graf, vector<int> &ways) {
    ways[v] = -1;
    for (auto el : graf[v]) {
        if (ways[el] != -1) {
            dfs3(el, graf, ways);
        }
    }
}

void dfs2(int v, vector<vector<int>> &graf,  vector<int> &ways) {
    ways[v] = 2;
    for (auto el : graf[v]) {
        if (ways[el] == 1) {
            dfs2(el, graf, ways);
        }
    }
}

void dfs(int v, vector<int> &vis, vector<int> &cycle, vector<vector<int>> &graf,  vector<int> &ways, vector<int> &progr) {
    if (progr[v] == 1) {
        cycle.push_back(v);
        ways[v] = -1;
    } else if (progr[v] == 2) {
        vis.push_back(v);
        ways[v] = 2;
    } else {
        progr[v] = 1;
        for (auto el : graf[v]) {
            dfs(el, vis, cycle, graf, ways, progr);
        }
        progr[v] = 2;
        ways[v] = 1;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        if (m == 0) {
            cout << 1 << " ";
            for (int i = 1; i != n; i++) {
                cout << 0 << " ";
            }
        } else {
            vector<vector<int>> graf(n);
            vector<int> ways(n, 0);
            vector<int> progr(n, 0);
            // 1 - in progress, 0 - new, 2 - visited
            vector<int> cycle;
            vector<int> vis;
            for (int i = 0; i != m; i++) {
                int s, f;
                cin >> s >> f;
                graf[s - 1].push_back(f - 1);
            }
            dfs(0, vis, cycle, graf, ways, progr);
            for (auto el : vis) {
                dfs2(el, graf, ways);
            }
            for (auto el : cycle) {
                dfs3(el, graf, ways);
            }
            for (auto el : ways) {
                cout << el << " ";
            }
        }
        cout << '\n';
    }
    return 0;
}