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;
typedef pair<int, int> pii;
typedef vector<int> vi;
#define all(v) v.begin(), v.end()
#define pb push_back    
#define sz(x) (int)(x).size()
void solve() { 
    int n, m;
    scanf("%d%d", &n, &m);
    vi vis(n, 0), clr(n, 1);
    vector<vi> g(n);
    for(int i = 0; i < m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        --u, --v;
        g[u].pb(v);
    }   
    // 1 ongoing
    // 2 done 
    function<void(int)> init_dfs = [&](int u) { 
        vis[u] = 1;
        for(int v : g[u]) { 
            if(!vis[v]) {
                init_dfs(v);
            } else if(vis[v] == 1) { 
                clr[v] = -1;
            } else if(clr[v] != -1){
                clr[v] = 2;
            }
        }
        vis[u] = 2;
    };
    function<void(int, int)> clr_dfs = [&](int u, int c) { 
        clr[u] = c;
        for(int v : g[u]) {
            if(clr[v] == 1 || (clr[v] == 2 && c == -1) ) {
                clr_dfs(v, c);
            }
        }
    };
    init_dfs(0);
    for(int u = 0; u < n; ++u) {
        if(clr[u] != 1) {
            clr_dfs(u, clr[u]);
        }
    }
    for(int i = 0; i < n; ++i) { 
        printf("%d ", vis[i] ? clr[i] : 0);
    }
    puts("");
}
int main() {
    int t = 1;
    scanf("%d",&t);
    while(t--)
        solve();
}