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"
using namespace std;
#define ch(i) ((i) + n  + 100)
#define MAXN 400050//点数
#define MAXM 400050//道路数
int n;

#define search(i,y) for(int i=head[y];i;i=edge[i].next) //遍历一个点的出边
#define add(u,v,c)  {edge[++cnt]={u,v,c,head[u]};head[u]=cnt;}//加入一条边
int head[MAXN],cnt=1;//0说明没有下一个值
struct EDGE{
    int u,v,c;
    int next;//下一个值的位置;
}edge[MAXM];
int dfn[MAXN]={0},low[MAXN];    //dfn记录深搜到的顺序    low记录一个点能向下追溯到的最大值
int num=0,code=0,grp[MAXN];
int selfroop[MAXN];
bool ins[MAXN];
stack<int> s;
int sz[MAXN];
void tarjan(int x) {
    dfn[x] = low[x] = ++num;
    s.push(x), ins[x] = 1;
    search(i, x) {
        int y = edge[i].v;
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        } else if (ins[y])//如果这个追溯回去的节点不在队列里面就不可能存在强连通
            low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] <= low[x]) {
        code++;
        int k=s.top();
        s.pop();grp[k]=code,ins[k]=0;
        sz[code] += selfroop[k];
        while (x != k){
            k = s.top(), s.pop();
            ins[k] = 0;
            grp[k] = code;
            sz[code] += selfroop[k];
        }
    }
}
void Tarjan(){
    tarjan(1);
}
const int inf = 0x3f3f3f3f;
int ans[MAXN],tag[MAXN],used[MAXN];
void dfs(int x,int fa){
    if(tag[x] || ans[x] >= 2)return ;
    if(ans[x] == -1)ans[x] = 0;
    ans[x] += ans[fa];
    search(i,x){
        int y = edge[i].v;
        dfs(y,x);
    }
}
void dfs2(int x){
    if(used[x]) return ;
    used[x] = true;
    search(i,x){
        int y = edge[i].v;
        if(used[y])continue;
        tag[y] = 1;
        ans[y] = -1;
        dfs2(y);
    }
}
int main(){
    int tot;cin>>tot;
    while(tot--){
        int m;
        cin>>n>>m;
        for(int i = 0;i<=  n ;i++)
            dfn[i] = low[i] = grp[i] = ins[i] = head[i] =sz[i] = tag[i] = ans[i] = used[i] = 0,selfroop[i] = 1;
        while(!s.empty())s.pop();
        num = 0,code = 0,cnt = 1;
        for(int i = 1;i<=m;i++){
            int a,b;
            cin>>a>>b;
            if(a == b){
                selfroop[a] ++;
            }else
            add(a,b,1);
        }
        Tarjan();
        for(int i = 1;i<=n;i++){
            ans[i] = -1;
            if(sz[grp[i]] >1)
                tag[i] = 1;
            if(sz[grp[i]] == 0)
                ans[i] = 0;
        }
        for(int i = 1;i<=n;i++){
            if(tag[i])
                dfs2(i);
        }
        if(!tag[1]){
            ans[1] = 1;
            dfs(1,0);
        }
        for(int i = 1;i<=n;i++){
            cout<<min(ans[i],2)<<" ";
        }
        cout<<endl;



    }
}