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;

const int N = 400010;
int vis[N];
vector<int> adj[N];
int dfn[N];
int low[N];
int in[N];
int ans[N];
stack<int> st;
int n, m, cnt;
int _;

void init(){
    cnt = 0;
    while(!st.empty()){
        st.pop();
    }
    for(int i = 1; i <= n; i++){
        dfn[i] = low[i] = in[i] = ans[i] = 0;
        vis[i] = 0;
        adj[i].clear();
    }
}

void tarjan(int u){
    low[u] = dfn[u] = ++cnt;
    st.push(u);
    in[u] = 1;
    for(auto v : adj[u]){
        if(v == u){
            ans[u] = -1;
        }
        if(!dfn[v]){
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(in[v]){
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u]){
        if(st.top() == u){
            in[u] = 0;
            st.pop();
        }
        else{
            int v;
            do{
                v = st.top();
                st.pop();
                ans[v] = -1;
                in[v] = 0;
            }while(u != v);
        }
    }
}

void solve(int u, int fa){
    if(fa == -1)
        ans[u] = -1;
    
    if(ans[u] == -1){
        return ;
    }

    ans[u] = min(ans[u] + fa, 2);
}

void dfs(int u){
    if(vis[u]){
        return ;
    }
    vis[u] = 1;
    for(auto v : adj[u]){
        solve(v, ans[u]);
        dfs(v);
    }
}

void dfs2(int u){
    for(auto v : adj[u]){
        if(ans[v] == 1){
            ans[v] = 2;
            dfs2(v);
        }
    }
}

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

    cin >> _;

    while(_--){
        cin >> n >> m;
        init();
        for(int i = 1; i <= m; i++){
            int a, b;
            cin >> a >> b;
            adj[a].push_back(b);
            // adj[b].push_back(a);
        }

        tarjan(1);

        solve(1, 1);

        for(int i = 1; i <= n; i++){
            if(ans[i] == -1){
                dfs(i);
            }
        }

        dfs(1);

        for(int i = 1; i <= n; i++){
            if(ans[i] == 2){
                dfs2(i);
            }
        }
        for(int i = 1; i <= n; i++){
            cout << ans[i] << " ";
        }
        cout << "\n";
    }
    return 0;
}