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;

#define ar array
#define ll long long
typedef int uci;
#define int long long
#define F first
#define S second

const int MAX_N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;

//! function insert

//THINK FIRST, CODE SECOND
//DON'T GET STUCK ON ONE STRATEGY
//CALM DOWNNN FOR ONCE IN YOUR LIFE
//REDUCE
//COUGH E??!?!?!! O.O
//uwu i see you ryan

void mark(int v, set<int> &curr, vector<bool> &loop, vector<bool> &multi, vector<vector<int>> &adj, vector<bool> &visited){
    visited[v] = true;
    for(int i{}; i < adj[v].size(); ++i){
        if(curr.count(adj[v][i])){
            loop[adj[v][i]] = true;
        }else if(visited[adj[v][i]]){
            multi[adj[v][i]] = true;
        }else{
            curr.insert(adj[v][i]);
            mark(adj[v][i], curr, loop, multi, adj, visited);
            curr.erase(curr.find(adj[v][i]));
        }
    }
}

void setouts(int v, bool inloop, bool inmulti, vector<bool> &loop, vector<bool> &multi, vector<vector<int>> &adj, vector<bool> &visited, vector<int> &outs){
    visited[v] = true;
    if(inloop|loop[v]){
        outs[v] = -1;
    }else if(multi[v]|inmulti){
        outs[v] = 2;
    }else{
        outs[v] = 1;
    }
    for(int i{}; i < adj[v].size(); ++i){
        if(inloop|loop[v] && outs[adj[v][i]] != -1){
            setouts(adj[v][i], inloop|loop[v], inmulti|multi[v], loop, multi, adj, visited, outs);
        }else if(multi[v]|inmulti && outs[adj[v][i]] != 2 && outs[adj[v][i]] != -1){
            setouts(adj[v][i], inloop|loop[v], inmulti|multi[v], loop, multi, adj, visited, outs);
        }else if(!visited[adj[v][i]]){
            setouts(adj[v][i], inloop|loop[v], inmulti|multi[v], loop, multi, adj, visited, outs);
        }
    }
}

void solve() {
    int n, m;
    cin >> n >> m;
    vector<bool> loop(n+1), multi(n+1), visited(n+1);
    vector<vector<int>> adj(n+1);
    for(int i{}; i < m; ++i){
        int a, b;
        cin >> a >> b;

        adj[a].push_back(b);
    }
    set<int> current{1};
    mark(1, current, loop, multi, adj, visited);
    vector<int> outs(n+1);
    visited.clear();
    visited.resize(n+1);
    setouts(1, loop[1], multi[1], loop, multi, adj, visited, outs);
    for(int i{1}; i <= n; ++i)
        cout << outs[i] << ' ';
    cout << '\n';
}

uci main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int tc; cin >> tc;
    for (int t = 1; t <= tc; t++) {
        // cout << "Case #" << t  << ": ";
        solve();
    }
}