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 <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <string>
#include <sstream>
#include <iomanip>
#include <map>
#include <unordered_map>
#include <stack>
#include <cstdio>
#include <climits>
#include <tuple>
#include <ctime>
#include <cstring>
#include <numeric>
#include <functional>
#include <chrono>
#include <cassert>
#include <bitset>
#define itn int
#define sacnf scanf
// printf("%.10f\n", ans);
using ll = long long;
using namespace std;
const ll mod = 1e9 + 7;
const int N = 1e5 + 1;
int n, m;
vector<vector<int>> gp;
vector<int> ans, cnt;
vector<bool> vis, cycles, path;

void dfs1(int u)
{
    vis[u] = 1;
    for (int& v : gp[u]) if (!vis[v]) dfs1(v);
}

void dfs2(int u)
{
    vis[u] = 1;
    path[u] = 1;
    for (int& v : gp[u])
    {
        if (path[v]) cycles[v] = 1;
        if (!vis[v]) dfs2(v);
    }
    path[u] = 0;
}

void dfs3(int u)
{
    vis[u] = 1;
    for (int& v : gp[u]) if (!vis[v]) dfs3(v);
}

void dfs4(int u)
{
    vis[u] = 1;
    for (int& v : gp[u])
    {
        cnt[v]++;
        if (!vis[v]) dfs4(v);
    }
}

void dfs5(int u)
{
    ans[u] = 2;
    vis[u] = 1;
    for (int& v : gp[u]) if (!vis[v] && ans[v] == -2) dfs5(v);
}

void test_case()
{
    scanf("%d%d", &n, &m);
    gp = vector<vector<int>>(n);
    ans = vector<int>(n, -2);
    cnt = vector<int>(n);
    vis = cycles = path = vector<bool>(n);
    for (int i = 0; i < m; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        gp[--u].push_back(--v);
    }
    dfs1(0);
    for (int i = 0; i < n; i++) if (!vis[i]) ans[i] = 0;
    vis = vector<bool>(n);
    dfs2(0);
    vis = vector<bool>(n);
    for (int i = 0; i < n; i++) if (cycles[i] && !vis[i]) dfs3(i);
    for (int i = 0; i < n; i++) if (vis[i]) ans[i] = -1;
    vis = vector<bool>(n);
    dfs4(0);
    vis = vector<bool>(n);
    for (int i = 0; i < n; i++) if (!vis[i] && cnt[i] >= 2 && ans[i] == -2) dfs5(i);
    for (int i = 0; i < n; i++) if (ans[i] == -2) ans[i] = 1;
    for (int& i : ans) printf("%d ", i);
    printf("\n");
}

int main()
{
    /*ios_base::sync_with_stdio(false);
    cin.tie(NULL);*/
    int tc;
    scanf("%d", &tc);
    while (tc--)
        test_case();

    return 0;
}