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:
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#include <unordered_map>

using namespace std;
typedef long long ll;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> g(n), gr(n);
    vector<int> p(n);

    for (int i = 0; i < m; i++) {
        int f, t;
        cin >> f >> t;
        f--;
        t--;
        g[f].push_back(t);
        gr[t].push_back(f);
        if (f == t)
            p[f] = 1;
    }

    vector<int> ord;
    vector<int> used(n);

    function<void(int)> dfs1 = [&](int v) {
        used[v] = 1;
        for (int to : g[v]) {
            if (!used[to])
                dfs1(to);
        }
        ord.push_back(v);
    };

    for (int i = 0; i < n; i++) {
        if (!used[i])
            dfs1(i);
    }

    fill(used.begin(), used.end(), 0);
    vector<int> comp;
    vector<int> asgn(n, -1);

    function<void(int)> dfs2 = [&](int v) {
        used[v] = 1;
        for (int to : gr[v]) {
            if (!used[to])
                dfs2(to);
        }
        comp.push_back(v);
    };

    int c = 0;
    vector<int> r;

    for (int i = n - 1; i >= 0; i--) {
        int v = ord[i];
        if (!used[v]) {
            comp.clear();
            dfs2(v);
            for (int x : comp)
                asgn[x] = c;
            if (comp.size() > 1 || p[comp[0]])
                r.push_back(0);
            else
                r.push_back(1);
            c++;
        }
    }

    ord.clear();
    fill(used.begin(), used.end(), 0);

    vector<vector<int>> g2(c), gr2(c);

    for (int i = 0; i < n; i++) {
        for (int to : g[i]) {
            if (asgn[i] != asgn[to]) {
                g2[asgn[i]].push_back(asgn[to]);
                gr2[asgn[to]].push_back(asgn[i]);
            }
        }
    }

    function<void(int)> dfs3 = [&](int v) {
        used[v] = 1;
        for (int to : g2[v]) {
            if (!used[to])
                dfs3(to);
        }
        ord.push_back(v);
    };

    for (int i = 0; i < c; i++) {
        if (!used[i]) {
            dfs3(i);
        }
    }

    reverse(ord.begin(), ord.end());

    vector<int> r2(c);
    vector<int> res(n);
    vector<int> r3(c);

    for (int v : ord) {
        bool inf = 0;
        int cnt = 0;

        if (asgn[0] == v) {
            cnt = 1;
        }

        for (int to : gr2[v]) {
            if (r2[to] == -1)
                inf = 1;
            cnt += r3[to];
        }

        if (cnt > 1)
            r3[v] = 2;
        else
            r3[v] = cnt;

        inf |= !r[v];

        if (r3[v] > 0 && inf)
            r2[v] = -1;
        else if (cnt > 1)
            r2[v] = r3[v];
        else
            r2[v] = r3[v];
    }

    for (int i = 0; i < n; i++) {
        res[i] = r2[asgn[i]];
    }

    for (int x : res)
        cout << x << ' ';
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#if defined(_DEBUG)
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int q = 1;
    cin >> q;
    for (; q > 0; q--) {
        solve();
        cout << '\n';
    }
}