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>
//#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;


typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;

#define pb push_back
#define eb emplace_back
#define ff first
#define ss second
#define REP(i, a, b) for(int i = a; i < b; i++)
#define INF int(2e9 + 9)
#define BIG_INF ll(9e18)
#define SIZE 400010
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define len(v) ll((v).size())
#define vec vector
#define sqr(x) (x) * (x)
#define cube(x) (x) * (x) * (x)
#define debug(x) cerr<<"[ "<<(#x)<<": "<<x<<" ] "


//
//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")


ll binpow(ll a, ll b, ll MOD = BIG_INF)
{
    ll res = 1;
    a %= MOD;
    while(b)
    {
        if(b & 1)
            res *= a;
        a *= a;
        b >>= 1;
        a %= MOD;
        res %= MOD;
    }

    return res % MOD;
}
template<typename T> void umax(T& a, T b){a = max(a, b);}
template<typename T> void umin(T& a, T b){a = min(a, b);}
clock_t startTime;
double getTime() {return (double)(clock() - startTime) / CLOCKS_PER_SEC;}


template<typename T>
ostream& operator<<(ostream& s, vector<T>& v) {
    for (auto el : v) {
        s << el << ' ';
    }
    return s;
}

ll mod = 998244353;

int T = 1;


int n, k, m;
vec<int> g[SIZE];
int used[SIZE];
bool isC[SIZE];
ll cnt[SIZE];
stack<int> path;


void dfs(int u) {

    used[u] = 1;
    path.push(u);
    for(int v: g[u]) {
        if(used[v] == 1) {
            stack<int> cur = path;
            while(cur.top() != v) {
                isC[cur.top()] = 1;
                cur.pop();
            }
            isC[cur.top()] = 1;
            continue;
        }
        if(used[v] == 2) continue;
        dfs(v);
    }
    used[u] = 2;
    path.pop();
}

void dfs2(int u, int p = -1) {

//    cout << u << '\n';
    used[u] = 1;
    for(int v: g[u]) {
        cnt[v] += cnt[u];
        if(isC[v] == 0 && isC[u] == 1) {
            isC[v] = 1;
            dfs2(v);
        }
        if(used[v]) continue;
        dfs2(v);
    }
    used[u] = 2;
}

void dfs3(int u) {
    used[u] = 1;
    for(int v: g[u]) {
        if(used[v]) continue;
        cnt[v] += cnt[u];
        dfs3(v);
    }
}



void solve() {

    cin >> n >> m;
    for(int i = 0; i < n; i++)
        g[i].clear();
    REP(i, 0, n) {
        used[i] = 0;
        isC[i] = 0;
        cnt[i] = 0;
    }
    REP(i, 0, m) {
        int u, v;
        cin >> u >> v;
        --u; --v;
        g[u].pb(v);
    }

    dfs(0);
    REP(i, 0, n) {
        used[i] = 0;
    }
    cnt[0] = 1;
    dfs2(0);
    REP(i, 0, n) {
        used[i] = 0;
    }
    REP(i, 0, n) {
        if(cnt[i] >= 2 && !used[i]) {
            dfs3(i);
        }
    }

    for(int i = 0; i < n; i++) {
        if(isC[i]) {
            cout << "-1 ";
            continue;
        }
        cout << min(cnt[i], 2ll) << ' ';
    }

    while(len(path)) path.pop();

    return;

}

int main()
{
    startTime = clock();
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    #ifdef LOCAL_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

    cin >> T;


    while(T--) {
        solve();
        cout << "\n";
    };
//    cout << getTime();
    return 0;
}