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:
//@Author- sitanshu
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define mii map<int, int>
#define umii unordered_map<int, int>
#define vi vector<int>
#define vl vector<ll>
#define clr(val) memset(val, 0, sizeof(val))
const ll MAX = 1000009, MOD = 1e9 + 7;
#define OJ                            \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);

vector<vector<int>> adj;
vector<int> st;
vector<int> vis;
vector<int> vis1;
vector<int> vis2;
void dfs(int s)
{
    vis[s] = 1;
    //st.pb(1);

    for (auto i : adj[s])
    {
        if (vis[i] == 1 || vis[i] == -1)
        {
            vis[i] = -1;
        }
        else if (vis[i] == 2)
        {
            vis[i] = 3;
        }
        else if (vis[i] == 0)
        {
            dfs(i);
        }
    }
    if (vis[s] == 1)
        vis[s] = 2;
}

void dfs1(int s)
{
    vis1[s] = 1;
    vis[s] = -1;
    // cout << s << " " << vis[s] << endl;
    for (auto i : adj[s])
    {

        if (vis1[i] == 0)
        {
            // cout << i << endl;
            dfs1(i);
        }
    }
}

void dfs2(int s)
{
    vis2[s] = 1;
    vis[s] = 3;

    for (auto i : adj[s])
    {
        if (vis[i] != -1 && vis[i] != 3 && vis2[i] == 0)
            dfs2(i);
    }
}

void solve()
{
    int n, m;
    cin >> n >> m;
    st.clear();
    adj = vector<vector<int>>(n + 1);
    vis = vector<int>(n + 1, 0);
    int x, y;
    for (int i = 0; i < m; i++)
    {
        cin >> x >> y;
        adj[x].pb(y);
    }
    vis1 = vector<int>(n + 1, 0);
    dfs(1);
    // for (int i = 1; i <= n; i++)
    //     cout << vis[i] << " ";
    // cout << endl;
    for (int i = 1; i <= n; i++)
    {
        if (vis[i] == -1 && vis1[i] == 0)
        {
            dfs1(i);
        }
    }
    // for (int i = 1; i <= n; i++)
    //     cout << vis[i] << " ";
    // cout << endl;
    vis2 = vector<int>(n + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        if (vis[i] == 3 && vis2[i] == 0)
            dfs2(i);
    }

    for (int i = 1; i <= n; i++)
    {
        if (vis[i] == 0)
            cout << 0 << " ";
        else if (vis[i] == 2)
            cout << 1 << " ";
        else if (vis[i] == 3)
            cout << 2 << " ";
        else
            cout << -1 << " ";
    }
    cout << endl;
}

int main()
{
    //  OJ;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}