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("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
*/
#include<bits/stdc++.h>
using namespace std;


#ifndef ONLINE_JUDGE
  #include "prettyprint.hpp"
#else 
#define dbg(...)
#endif
 
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
 
#define ar array 
#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define sz(v) int((v).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i,a,b) for(int i = a; i < b; i++)
#define deb(x) cout << #x << "=" << x << endl
#define PI 3.1415926535897932384626
#define PQ(type) priority_queue<type>
#define PQD(type) priority_queue<type,vector<type>,greater<type> >
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
#define endl '\n'
 
//typedef tree<int,null_type,less<int>,rb_tree_tag, tree_order_statistics_node_update> indexed_set; // use less_equal for multiset
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the iterator to kth largest element.(0-based)
 
typedef long long ll; 
typedef long double ld; 
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
 
 
const  int mxN = 1e5+5;
const  long long INF= 1e18;
const long long mod= 998244353;

mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define SHUF(v) shuffle(all(v), RNG);
// Use mt19937_64 for 64 bit random numbers.



void solve() {  
  int n, m;
  cin >> n >> m;

  vi vis(n + 1), instack(n + 1);

  vector<vi> adj(n + 1);

  rep(i, 0, m) {
    int a, b;
    cin >> a >> b;
    adj[a].pb(b);
  }

  vi group_a(n + 1), group_b(n + 1), ans(n + 1);

  function<void(int, int, int val)> dfs = [&] (int u, int mode, int val) -> void {
    vis[u] = 1;
    ans[u] = val;
    instack[u] = 1;
    for(auto x: adj[u]) {
      if(mode && vis[x]) {
        if(instack[x]) group_a[x] = 1;
        else group_b[x] = 1;
      }      
      if(!vis[x]) dfs(x, mode, val);
    }
    instack[u] = 0;
  };

  dfs(1, -1, 1);
  rep(i, 1, n + 1) vis[i] = 0;
  rep(i, 1, n + 1) {
    if(group_b[i] && !vis[i]) dfs(i, 0, 2);
  }
  rep(i, 1, n + 1) vis[i] = 0;
  rep(i, 1, n + 1) {
    if(group_a[i] && !vis[i]) dfs(i, 0, -1);
  }
  
  rep(i, 1, n + 1) {
    if(ans[i] == 3) ans[i] = -1;
    cout << ans[i] << " ";
  }
  cout << endl;

}

int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);

int t = 1;
cin >> t;
while(t--) {
solve();
}
}