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:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define all(arr) (arr).begin(), (arr).end()
#define ll long long
#define ld long double
#define pb push_back
#define sz(x) int((x).size())
#define fi first
#define se second
#define endl '\n'

const int N = 4e5 + 10;

vector<int> g[N], tg[N], newg[N];

int used[N], bad[N], ans[N], self[N], comp[N], cycle[N], compans[N];

void dfs1(int v){
  used[v] = 1;
  for (auto u : g[v]){
    if (used[u]) bad[u] = 1;
    else dfs1(u);
  }
}

void dfs2(int v){
  used[v] = 1;
  bad[v] = 1;
  for (auto u : g[v]){
    if (!used[u]) dfs2(u);
  }
}

vector<int> topsort;

void dfs3(int v){
  used[v] = 1;
  for (auto u : g[v]){
    if (!used[u]) dfs3(u);
  }
  topsort.pb(v);
}

int cnt = 0;

void dfs4(int v, int cnum){
  used[v] = 1;
  cnt++;
  comp[v] = cnum;
  for (auto u : tg[v]){
    if (!used[u]) dfs4(u, cnum);
  }
}

void dfs5(int v){
  used[v] = 1;
  compans[v] = -1;
  for (auto u : newg[v]){
    if (!used[u]) dfs5(u);
  }
}

void solve(){
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++){
    int a, b;
    cin >> a >> b;
    g[a].pb(b);
    tg[b].pb(a);
    if (a == b) self[a] = 1;
  }
  dfs1(1);
  vector<int> badv, reached(n + 10), rep(n + 10);
  for (int i = 1; i <= n; i++){
    if (used[i]) reached[i] = 1;
    if (bad[i]) badv.pb(i);
  }
  for (int i = 0; i <= n; i++) used[i] = 0;
  for (auto u : badv){
    if (!used[u]) dfs2(u);
  }
  for (int i = 0; i <= n; i++) used[i] = 0;
  for (int i = 1; i <= n; i++){
    if (!used[i]) dfs3(i);
  }
  reverse(all(topsort));
  for (int i = 0; i <= n; i++) used[i] = 0;
  int cnum = 1;
  for (auto i : topsort){
    if (!used[i]){
      cnt = 0;
      dfs4(i, cnum);
      rep[cnum] = i;
      if (cnt > 1 || self[i]) cycle[cnum] = 1;
      cnum++;
    }
  }
  for (int i = 1; i <= n; i++){
    for (auto u : g[i]) newg[comp[i]].pb(comp[u]);
  }
  for (int i = 0; i <= n; i++) used[i] = 0;
  for (int i = 1; i <= n; i++){
    if (cycle[i] && !used[i] && reached[rep[i]]) dfs5(i);
  }
  for (int i = 1; i <= n; i++){
    if (!reached[i]) continue;
    if (compans[comp[i]] == -1){
      ans[i] = -1;
    }
    else{
      if (bad[i]) ans[i] = 2;
      else ans[i] = 1;
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  topsort.clear(); cnt = 0;
  for (int i = 1; i <= n; i++){
    g[i].clear();
    tg[i].clear();
    newg[i].clear();
    used[i] = bad[i] = ans[i] = self[i] = comp[i] = cycle[i] = compans[i] = 0;
  }
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  // cout.precision(20);
  int t;
  cin >> t;
  while (t--){
    solve();
  }
}