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:
// Problem: G. How Many Paths?
// Contest: Codeforces - Codeforces Round #731 (Div. 3)
// URL: https://codeforces.com/contest/1547/problem/G
// Memory Limit: 512 MB
// Time Limit: 4000 ms
// Author:  UpMing
//
// Powered by CP Editor (https://cpeditor.org)

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef unsigned long long ull;

const long long INF = 1e18;
const int maxn = 4e5 + 7;
const ll mod = 1e9 + 7;

#define pb push_back
#define debug(x) cout << #x << ":" << x << endl;
#define mst(x, a) memset(x, a, sizeof(x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define dep(i, a, b) for (int i = (a); i >= (b); --i)

inline ll read() {
    ll x = 0;
    bool f = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        f |= ch == '-', ch = getchar();
    while ('0' <= ch && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
}

void out(ll x) {
    int stackk[20];
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (!x) {
        putchar('0');
        return;
    }
    int top = 0;
    while (x)
        stackk[++top] = x % 10, x /= 10;
    while (top)
        putchar(stackk[top--] + '0');
}
ll qpow(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1)
            ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
ll n, m, in[maxn], f[maxn],tpd[maxn];
vector<ll> e[maxn];
ll ans[maxn], dist[maxn];
void bfs() {
    queue<int> q;
    dist[1] = 1;
    q.push(1);
    while (q.size()) {
        int u = q.front();
        q.pop();
        for (int v : e[u]) {
            if (dist[v] == 0) {
                dist[v] = 1;
                q.push(v);
            }
        }
    }
}

void top_sort() {
	rep(i,1,n) ans[i] = dist[i];
	for(int i=1 ;i<=n;i++)
	{
		if(ans[i])
		{
			for(int j:e[i])
			{
				in[j]++;
			}
		}
	}
	queue<int>q;
	for(int i=1 ;i<=n ;i++) if(ans[i]&&in[i]==0) q.push(i),f[i] = 1;
	while(q.size())
	{
		int u = q.front();
		q.pop();
		tpd[u] = 1;
		for(int v:e[u])
		{
			in[v]--;
			f[v]+=f[u];
			f[v] = min(f[v],INF);
			if(in[v]==0) q.push(v);
		}
	}

	rep(i,1,n)
	{
		if(ans[i]){
			
			 if(tpd[i]==0) ans[i] =-1;
			 else if(f[i]>=2) ans[i] =2 ;
			//. else ans[i] = 1;
		}
	}
	
	rep(i,1,n)cout<<ans[i]<<" ";
	cout<<endl;
}

int main() {
    // ios::sync_with_stdio(false);
    int _ = read();
    while (_--) {
        n = read(), m = read();
        rep(i, 1, n) e[i].clear(),tpd[i] =  dist[i] = ans[i] = f[i] = in[i] = 0;
		
        rep(i, 1, m) {
            ll u, v;
            u = read(), v = read();
            e[u].pb(v);
        }
        bfs();
        top_sort();
    }
    return 0;
}
/*


*/