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>

class graph {
    int n;
    std::vector<std::vector<int>> E;

    int idx;
    std::vector<int> dfn, low;
    std::vector<int> sta;
    std::vector<bool> vis;

    int cnt;
    std::vector<int> bel;

    void tarjan(int u) {
        dfn[u] = low[u] = idx++;
        sta.push_back(u);
        vis[u] = true;
        for (int v : E[u]) {
            if (dfn[v] == -1) {
                tarjan(v);
                low[u] = std::min(low[u], low[v]);
            } else if (vis[v]) {
                low[u] = std::min(low[u], dfn[v]);
            }
        }
        if (dfn[u] == low[u]) {
            int v;
            do {
                v = sta.back();
                vis[v] = false;
                bel[v] = cnt;
                sta.pop_back();
            } while (v != u);
            ++cnt;
        }
    }

    std::vector<std::vector<int>> G;
    std::vector<int> deg;
    std::vector<int> ec;

public:
    graph(const std::vector<std::vector<int>> &_E)
    : n(_E.size()), E(_E), idx(0), dfn(n, -1), low(n), vis(n), cnt(0), bel(n) {
        for (int i = 0; i < n; ++i) {
            if (dfn[i] == -1) {
                tarjan(i);
            }
        }
        G.resize(cnt);
        ec.resize(cnt);
        deg.resize(cnt);
        for (int u = 0; u < n; ++u) {
            for (int v : E[u]) {
                if (bel[u] != bel[v]) {
                    G[bel[u]].push_back(bel[v]);
                    ++deg[bel[v]];
                } else {
                    ++ec[bel[u]];
                }
            }
        }
    }

    std::vector<int> get_ans(int s) const {
        s = bel[s];
        std::vector<int> f(cnt);
        f[s] = ec[s] ? -1 : 1;
        std::vector<int> d(deg);
        std::vector<int> Q;
        for (int i = 0; i < cnt; ++i) {
            if (!d[i]) {
                Q.push_back(i);
            }
        }
        for (int i = 0; i < (int)Q.size(); ++i) {
            int u = Q[i];
            for (int v : G[u]) {
                --d[v];
                if (f[u]) {
                    if (f[u] == -1 || f[v] == -1 || ec[v]) {
                        f[v] = -1;
                    } else {
                        f[v] = std::min(f[u] + f[v], 2);
                    }
                }
                if (!d[v]) {
                    Q.push_back(v);
                }
            }
        }
        std::vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            ans[i] = f[bel[i]];
        }
        return ans;
    }
};

void solve() {
    int n, m;
    std::cin >> n >> m;
    std::vector<std::vector<int>> E(n);
    for (int i = 0; i < m; ++i) {
        int u, v;
        std::cin >> u >> v;
        --u, --v;
        E[u].push_back(v);
    }
    graph G(E);
    auto ans = G.get_ans(0);
    for (int i = 0; i < n; ++i) {
        std::cout << ans[i] << " \n"[i == n - 1];
    }
}

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);

    int T;
    std::cin >> T;
    while (T--) {
        solve();
    }
}