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;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second
#define cl       clear
#define sep      ' '
#define endll    '\n'

const int maxn  =  1e6   + 10;
const int maxn5 =  5e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const ll  mod   =  1e9   +  7;
const ll  inf   =  2e18;

vector <int> srt, adj[maxn5], jda[maxn5], exadj[maxn5];
int val[maxn5], cmp[maxn5], cnt[maxn5], num = 0;
bool mark[maxn5];

void dfs1(int v){
    mark[v] = true;
    for(auto u : adj[v]) if(!mark[u]) dfs1(u);
    srt.pb(v);
    return;
}
void dfs2(int v){
    mark[v] = true;
    cmp[v] = num;
    cnt[num]++;
    //cout << v << ' ' << num << endl;
    for(auto u : jda[v]) if(!mark[u]) dfs2(u);
    return;
}
void dfs3(int v){
    mark[v] = true;
    for(auto u : exadj[v]) if(!mark[u]) dfs3(u);
    srt.pb(v);
    return;
}
int main()
{
    //cout << setprecision(15);
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    int tt; cin >> tt;
    while(tt--){
        int n, m; cin >> n >> m;
        for(int i = 0; i <= n; i++){
            adj[i].cl(); jda[i].cl(); exadj[i].cl();
        }
        for(int i = 0; i < m; i++){
            int a, b; cin >> a >> b;
            adj[a].pb(b);
            jda[b].pb(a);
        }
        srt.cl();
        fill(mark, mark + n + 10, false);
        fill(cnt, cnt + n + 10, 0);
        for(int i = 1; i <= n; i++) if(!mark[i]) dfs1(i);
        reverse(all(srt));
        fill(mark, mark + n + 10, false);
        num = 0;
        for(auto u : srt) if(!mark[u]){
            dfs2(u);
            num++;
        }
        for(int i = 1; i <= n; i++) for(auto u : adj[i]){
            if(cmp[i] != cmp[u]) exadj[cmp[i]].pb(cmp[u]);
            else if(u == i) cnt[cmp[i]]++;
        }
        fill(val, val + n + 10, 0);
        srt.cl();
        fill(mark, mark + n + 10, false);
        for(int i = 0; i < num; i++) if(!mark[i]) dfs3(i);
        reverse(all(srt));
        val[cmp[1]] = 1;
        if(cnt[cmp[1]] > 1) val[cmp[1]] = 3;
        for(auto v : srt){
            if(val[v] == 0) continue;
            if(cnt[v] > 1) val[v] = 3;
            for(auto u : exadj[v]){
                if(val[v] >= 2) val[u] = max(val[v], val[u]);
                else if(val[v] == 1) val[u] = max(val[u], min(2, val[u] + 1));
            }
        }
        for(int i = 1; i <= n; i++) cout << (val[cmp[i]] == 3 ? -1 : val[cmp[i]]) << ' ' ;
        cout << endll;
    }
    return 0;
}