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 long double ld;
typedef unsigned long long ull;
const ll MOD = 1e9 + 7;
const int iinf = 1 << 29;
const long long llinf = 1ll << 60;
const double PI = 3.14159265;
#define pb push_back
#define ff first
#define ss second
#define sz(a) ((int)(a).size())
#define all(x) (x).begin(),(x).end()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll mrand(ll B) {
  return (ull)rng() % B;
}

inline long long ceil(long long x, long long y) {
    return (x + y - 1LL) / y;
}

void dfs(vector<vector<int>> &g, vector<bool> &has_self, vector<int> &status, int node, int val, vector<bool> &in_path) {
    if (val <= status[node]) return;
    if (has_self[node]) val = 3;
    status[node] = val;
    in_path[node] = true;
    for (auto v : g[node]) {
        if (val == 3 || in_path[v]) {
            if (status[v] != 3) dfs(g, has_self, status, v, 3, in_path);
            continue;
        }

        if (status[v] == 0 || status[v] == 1) dfs(g, has_self, status, v, status[v] + 1, in_path);
    }
    in_path[node] = false;
}

void work() {
    int n, m; cin >> n >> m;
    vector<vector<int>> g(n);
    vector<bool> has_self(n, false);
    vector<int> status(n, 0);
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v; u--; v--;
        if (u == v) {
            has_self[u] = true;
        } else {
            g[u].pb(v);
        }
    }
    vector<bool> in_path(n, false);
    dfs(g, has_self, status, 0, 1, in_path);
    for (int i = 0; i < n; i++) cout << (status[i] == 3 ? -1 : status[i]) << ' ';
    cout << '\n';
}

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int nt; cin>>nt; while (nt--) work();
    //work();
    return 0;
}