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 <cstdio>
#include <cstring>

#define ll long long

int t, n, m;

int head[400005];
struct e {
    int to, next;
} edge[400005];
void addE(int x, int y) {
    edge[++head[0]] = {y, head[x]};
    head[x] = head[0];
}

int mk[400005], ls1[400005], ls2[400005], mk1[400005], mk2[400005];
void dfs1(int u) {
    mk[u] = 1;
    for (int i = head[u]; i; i = edge[i].next)
        if (mk[edge[i].to] == 0)
            dfs1(edge[i].to);
        else if (mk[edge[i].to] == 1)
            ls1[++ls1[0]] = edge[i].to;
        else
            ls2[++ls2[0]] = edge[i].to;
    mk[u] = 2;
}
void dfs2(int u) {
    mk1[u] = 1;
    for (int i = head[u]; i; i = edge[i].next)
        if (mk1[edge[i].to] == 0)
            dfs2(edge[i].to);
    mk1[u] = 2;
}
void dfs3(int u) {
    mk2[u] = 1;
    for (int i = head[u]; i; i = edge[i].next)
        if (mk2[edge[i].to] == 0)
            dfs3(edge[i].to);
    mk2[u] = 2;
}

int main() {
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        memset(head, 0, sizeof(int) * (n + 5));
        memset(mk, 0, sizeof(int) * (n + 5));
        memset(mk1, 0, sizeof(int) * (n + 5));
        memset(mk2, 0, sizeof(int) * (n + 5));
        ls1[0] = ls2[0] = 0;
        for (int i = 0, a, b; i < m; ++i) {
            scanf("%d%d", &a, &b);
            addE(a, b);
        }
        dfs1(1);
        for (int i = 1; i <= ls1[0]; ++i)
            dfs2(ls1[i]);
        for (int i = 1; i <= ls2[0]; ++i)
            dfs3(ls2[i]);
        for (int i = 1; i <= n; ++i)
            printf(" %d" + (i == 1), mk[i] ? (mk1[i] ? -1 : (mk2[i] ? 2 : 1)) : 0);
        putchar('\n');
    }
    return 0;
}