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;

const int N = 4e5 + 5;
const int M = 18;
const int OO = 1e9;
const int mod = 1e9 + 7;

vector< vector<int> > scc;
vector<int> adj[N], src;
int n, m, dfsn[N], low[N], timer, ans[N];
bool in_stack[N], self[N];
stack<int> st;

void tarjan(int node) {
    dfsn[node] = low[node] = ++timer;
    in_stack[node] = true;
    st.push(node);
    for (auto i: adj[node]) {
        if (!dfsn[i]) {
            tarjan(i);
            low[node] = min(low[node], low[i]);
        } else if (in_stack[i]) low[node] = min(low[node], dfsn[i]);
    }
    if (dfsn[node] == low[node]) {
        scc.emplace_back();
        while (true) {
            int cur = st.top();
            st.pop();
            in_stack[cur] = false;
            scc.back().push_back(cur);
            if (cur == node) break;
        }
    }
}

void dfs(int node, int val) {
    if(ans[node] == val || ans[node] == -1) {
        return;
    }
    ans[node] = val;
    for(auto &v: adj[node]) {
        dfs(v, val);
    }
}

void traverse(int node) {
    if(ans[node]) {
        if(ans[node] == 1) {
            ans[node] = 3;
            src.push_back(node);
        }
        return;
    }
    ans[node] = 1;
    for(auto &v : adj[node]) {
        traverse(v);
    }
}

void init() {
    timer = 0;
    memset(ans, 0, (n + 1) * sizeof ans[0]);
    memset(dfsn, 0, (n + 1) * sizeof dfsn[0]);
    memset(low, 0, (n + 1) * sizeof low[0]);
    memset(self, 0, (n + 1) * sizeof self[0]);
    scc.clear();
    for (int i = 1; i <= n; ++i) {
        adj[i].clear();
    }
    src.clear();
}

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif

    int tt;
    cin >> tt;
    while (tt--) {
        cin >> n >> m;
        init();
        while(m--){
            int u, v;
            cin >> u >> v;
            adj[u].push_back(v);
            if(u == v) {
                self[u] = true;
            }
        }
        tarjan(1);
        for(auto &comp : scc) {
            if(comp.size() == 1 && !self[comp.front()]) {
                continue;
            }
            for(auto &node : comp) {
                dfs(node, -1);
            }
        }
        if(ans[1] != -1) {
            traverse(1);
            for(auto &node: src) {
                dfs(node, 2);
            }
        }
        for (int i = 1; i <= n; ++i) {
            cout << ans[i] << ' ';
        }
        cout << '\n';
    }
    return 0;
}