Problem: There are n segments [l_i, r_i] for 1 ≤ i ≤ n. You should divide all segments into two non-empty groups in such way that there is no pair of segments from different groups which have at least one common point, or say that it's impossible to do it. Each segment should belong to exactly one group.

To optimize testing process you will be given multitest.

Input

The first line contains one integer T (1 ≤ T ≤ 50000) — the number of queries. Each query contains description of the set of segments. Queries are independent.

First line of each query contains single integer n (2 ≤ n ≤ 10^5) — number of segments. It is guaranteed that ∑{n} over all queries does not exceed 10^5.

The next n lines contains two integers l_i, r_i per line (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the i-th segment.

Output

For each query print n integers t_1, t_2, ..., t_n (t_i ∈ \{1, 2\}) — for each segment (in the same order as in the input) t_i equals 1 if the i-th segment will belongs to the first group and 2 otherwise.

If there are multiple answers, you can print any of them. If there is no answer, print -1.

Example

Input


3
2
5 5
2 3
3
3 5
2 3
2 3
3
3 3
4 4
5 5


Output


2 1 
-1
1 1 2 

Note

In the first query the first and the second segments should be in different groups, but exact numbers don't matter.

In the second query the third segment intersects with the first and the second segments, so they should be in the same group, but then the other group becomes empty, so answer is -1.

In the third query we can distribute segments in any way that makes groups non-empty, so any answer of 6 possible is correct.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class CF1101C_DivisionAndUnion {
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		public Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

	}

	static void print(Object b) {
		System.out.print(b);
	}

	static void println(Object b) {
		System.out.println(b);
	}

	static void println() {
		System.out.println();
	}

	static class Node implements Comparable<Node> {
		int val;
		int id;
		int type; // 1 is start 2 is end

		public Node(int v, int i, int t) {
			val = v;
			id = i;
			type = t;
		}

		@Override
		public int compareTo(Node o) {
			if (val == o.val)
				return type - o.type;
			return val - o.val;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner();
		int test = scanner.nextInt();
		PriorityQueue<Node> q = new PriorityQueue<Node>();
		for (int te = 0; te < test; te++) {
			int n = scanner.nextInt();
			for (int i = 0; i < n; i++) {
				q.add(new Node(scanner.nextInt() - 1, i, 1));
				q.add(new Node(scanner.nextInt() - 1, i, 2));
			}
			int count = 0;
			int[] ans = new int[n];
			int type = 1;
			for (int i = 0; i < 2 * n; i++) {
				Node node = q.poll();
				if (node.type == 1)
					count++;
				else {
					count--;
					ans[node.id] = type;
					if (count == 0 && i != 2*n-1) {
						type = 2;
					}
				}
			}
			if (type == 1)
				println(-1);
			else {
				for (int i = 0; i < n; i++)
					print(ans[i] + " ");
			println();
			}
		}
	}

}