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:
                        //    I solemnly swear that I am upto no good //

#include <bits/stdc++.h>
using namespace std;
#define sub             freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);
#define ll              long long
#define ull             unsigned long long
#define ld              long double
#define ttime           {cerr << '\n'<<"Time (in s): " << double(clock() - clk) * 1.0 / CLOCKS_PER_SEC << '\n';}
#define helpUs          template<typename T = ll , typename U = ll>
#define helpMe          template<typename T = ll>
#define pb              push_back
#define sz(x)           (int)((x).size())
#define fast            ios_base::sync_with_stdio(false);cin.tie(0);
#define all(x)          (x).begin(),(x).end()
#define rep(i,a,b)      for(ll i=a;i<b;i++)
#define pr(x)           cout << #x " = " << (x) << "\n"
#define mp              make_pair
#define ff              first
#define ss              second
#define YY              cout<<"Yes"<<endl
#define NN              cout<<"No"<<endl
#define ppc             __builtin_popcount
#define ppcll           __builtin_popcountll

const long long INF=1e18;
const long long N=400005;
const long long mod=1000000007;                    // 998244353, 2971215073, 1000050131, 433494437

#define endl "\n"
#define int ll

typedef pair<ll,ll> pairll;
typedef map<ll,ll>  mapll;
typedef map<char,ll> mapch;
typedef vector<ll> vll;
mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
helpUs class comp{public:bool operator()(T a, U b){return a>b;}};
helpUs istream& operator>>(istream& aa, pair<T,U> &p){aa>>p.ff>>p.ss;return aa;}
helpMe ostream& operator<<(ostream& ja, vector<T> &v){for(auto it:v)ja<<it<<" ";return ja;}
helpMe istream& operator>>(istream& aa, vector<T> &v){for(auto &it:v)cin>>it;return aa;}
helpUs ostream& operator<<(ostream& ja, pair<T,U> &p){ja<<p.ff<<" "<<p.ss;return ja;}

ll n,k,tt=1;
int white=0, grey=1, black=2;
vll adj[N], color(N), visited(N), infinites, multiple,ans;

void dfs(int x){
    if(color[x]==grey){
        infinites.pb(x);return;
    }
    else if(color[x]==black){
        multiple.pb(x);
        return;
    }
    color[x]=grey;
    for(auto it:adj[x]){
    // cout<<it<<endl;
        
            dfs(it);
        
    }
    color[x]=black;
}

void dfsINF(int x){
    visited[x]=1;  
    ans[x]=-1;
    for(auto it:adj[x]){
        if(!visited[it]){
            dfsINF(it);
        }
    }

}

void dfsMul(int x){
    visited[x]=1;  
    ans[x]=2;
    for(auto it:adj[x]){
        if(!visited[it]){
            dfsMul(it);
        }
    }

}

void solve(){

    string s,t;
    ll m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)adj[i].clear(), color[i]=0, visited[i]=0;
    ans=vector<int> (n+1,0);
    infinites.clear();
    multiple.clear();
    for(int i=0;i<m;i++){
        ll a,b;
        cin>>a>>b;
        adj[a].pb(b);
    }

    dfs(1);

    for(auto it:infinites){
        if(!visited[it])
        dfsINF(it);
    }
    for(auto it:multiple){
        if(!visited[it])
        dfsMul(it);
    }

    for(int i=1;i<=n;i++){
        if(!visited[i] && (color[i]==black))
            ans[i]=1;
    }

    for(int i=1;i<=n;i++)
        cout<<ans[i]<<" ";

    cout<<endl;
    
    

}
           
signed main(){
    fast;
    ll t=1;
    #ifndef ONLINE_JUDGE
        sub;    
    #endif
    // clock_t clk = clock();
    // pre();

    cin>>t;
    while(t--)
        solve();
    // ttime;
    return 0;
}        

                                // Mischief Managed //