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;

#define ll long long
#define pi pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long>
#define vpi vector<pair<int, int>>
#define vvi vector<vector<int>>
#define mii map<int, int>
#define mmii multimap<int, int>
#define umii unordered_map<int, int>
#define umsii unordered_multimap<int, int>
#define si set<int>
#define msi multiset<int>
#define usi unordered_set<int>
#define umsi unordered_multiset<int>
#define sc set<char>

#define f(i, s, e) for (long long int i = s; i < e; i++)
#define cf(i, s, e) for (long long int i = s; i <= e; i++)
#define rf(i, e, s) for (long long int i = e - 1; i >= s; i--)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

template <class T>
void print_v(vector<T> &v) {
    cout << "[";
    for (auto x : v) cout << x << ", ";
    cout << "\b\b]\n";
}

template <class T>
long long int sum_v(vector<T> &v) {
    long long int ans = 0;
    for (auto a : v) ans += a;
    return ans;
}

#define MOD 1000000007
#define PI 3.1415926535897932384626433832795
ll min(ll a, int b) {
    if (a < b) return a;
    return b;
}
ll min(int a, ll b) {
    if (a < b) return a;
    return b;
}
ll max(ll a, int b) {
    if (a > b) return a;
    return b;
}
ll max(int a, ll b) {
    if (a > b) return a;
    return b;
}
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
string to_upper(string a) {
    for (int i = 0; i < (int)a.size(); ++i)
        if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A';
    return a;
}
string to_lower(string a) {
    for (int i = 0; i < (int)a.size(); ++i)
        if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A';
    return a;
}
bool prime(ll a) {
    if (a == 1) return 0;
    for (int i = 2; i <= round(sqrt(a)); ++i)
        if (a % i == 0) return 0;
    return 1;
}
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }

typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int uint64;

void dfs(vvi &graph, vi &ans, si &visited, int node) {
    if (ans[node] == -1 || ans[node] == 2) return;
    if (visited.count(node)) {
        vi stack;
        vi tempvisited(graph.size());
        stack.push_back(node);
        while (!stack.empty()) {
            int val = stack.back();
            stack.pop_back();
            ans[val] = -1;
            tempvisited[val] = 1;
            for (int i : graph[val]) {
                if (!tempvisited[i]) stack.push_back(i);
            }
        }
        return;
    }
    visited.insert(node);
    if (ans[node] != -1) ans[node] = min(ans[node] + 1, 2);
    for (int i : graph[node]) {
        dfs(graph, ans, visited, i);
    }
    visited.erase(node);
}

void solve() {
    int n, m;
    cin >> n >> m;
    vvi graph(n + 1);
    vi ans(n + 1);
    si visited;
    f(i, 0, m) {
        int a, b;
        cin >> a >> b;
        graph[a].push_back(b);
    }
    dfs(graph, ans, visited, 1);
    cf(i, 1, n) cout << ans[i] << " ";
    cout << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}