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 ll long long int
#define ii pair<ll,ll>
#define MP make_pair
#define MOD7 1000000007
#define MOD9 1000000009
void printGoogleAns(int ans, int caseNum){
    cout << "Case #" << caseNum << ": " << ans << endl;
}

// BLACK - 0
//WHITE - 1
//GREY - 2
ll t;
ll n, m;
vector<vector<int>> g;
set<ll> s[2];
vector<int> col;
void dfs(int node){
    col[node] = 2;
    for(auto v: g[node]){
        if(col[v] == 1){
            dfs(v);
        }else{
            if(col[v] == 0){
                s[0].insert(v);
            }else{
                s[1].insert(v);
            }
        }
    }
    col[node] = 0;
}

void normaldfs(int node, vector<int> &col1){
    //cout << "reached :" << node << endl;
    col1[node] = 0;
    for(auto v: g[node]){
        if(col1[v] == 1){
            //cout << "checking :" << v << " " << col1[v] << endl;
            normaldfs(v, col1);
        }
    }
}

void solve(){
    col.clear();
    for(auto &i: s){
        i.clear();
    }
    g.clear();
    cin >> n >> m;
    col.resize(n + 1, 1);
    g.resize(n + 1);
    for(int i = 0;i < m;i++){
        ll s,e;
        cin >> s >> e;
        g[s].push_back(e);
    }

    dfs(1);
    // cout << "2 paths :";
    // for(auto i: s[0]){
    //     cout << i << " ";
    // }
    // cout << endl;
    // cout << "infinite paths :";
    // for(auto i: s[1]){
    //     cout << i << " ";
    // }
    // cout << endl;
    // cout << "col : ";
    // for(auto i: col){
    //     cout << i << " ";
    // }
    // cout << endl;
    vector<int> col1;
    col1.clear();
    col1.resize(n + 1, 1);
    for(auto i: s[0]){
        if(col1[i] == 1){
            normaldfs(i, col1);
        }
    }
    // cout << "col1 : ";
    // for(auto i: col1){
    //     cout << i << " ";
    // }
    // cout << endl;
    vector<int> col2;
    col2.resize(n + 1, 1);
    for(auto i: s[1]){
        if(col2[i]== 1){
            normaldfs(i, col2);
        }
    }
    // cout << "col2 : ";
    // for(auto i: col2){
    //     cout << i << " ";
    // }
    // cout << endl;

    vector<int> ans(n + 1,0);
    for(int i = 1;i < n + 1;i++){
        if(col[i] == 1){
            ans[i] = 0;
        }else{
            if(col2[i] == 0){
                ans[i] = -1;
            }else if(col1[i] == 0){
                ans[i] = 2;
            }else{
                ans[i] = 1;
            }
        }
    }
    for(int i = 1;i < n + 1;i++){
        cout << ans[i] << " ";
    }
    cout << endl;
}

signed main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> t;
    while(t--)
    solve();
    return 0;
}