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.util.*;

public class CF1101C {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i = 0; i < t; i++){
            query(sc);
        }
    }

    private static void query(Scanner sc){
        int n = sc.nextInt();

        if(n == 1){
            System.out.println(-1);
            return;
        }

        //从输入取值
        ArrayList<Line> list = new ArrayList<Line>();
        for(int i = 0; i < n; i++){
            int l = sc.nextInt();
            int r = sc.nextInt();
            list.add(new Line(l,r));
        }

        //按照线段的左节点排序
        ArrayList<Line> sortedList = (ArrayList<Line>) list.clone();
        Collections.sort(sortedList,new ListComparator());

        //循环线段，寻找一个合适的集合分隔点
        boolean hasSplitPoint = false;
        int splitPoint = sortedList.get(0).getR();
        for(int i = 1; i < n; i++){
            Line line = sortedList.get(i);
            if(line.getL() > splitPoint){
                hasSplitPoint = true;
                break;
            }else{
                splitPoint = Math.max(splitPoint, line.getR());
            }
        }

        if(!hasSplitPoint){
            System.out.println(-1);
            return;
        }

        //根据分隔点，判断线段所在集合：在分隔点左边的线段是集合1，在分隔点右边的线段是集合2
        StringBuffer sb = new StringBuffer();
        for(int i = 0; i < n; i++){
            Line line = list.get(i);
            if(line.getR() <= splitPoint){
                sb.append("1 ");
            }else{
                sb.append("2 ");
            }
        }
        System.out.println(sb.toString());
    }


}

//排序比较器
class ListComparator implements Comparator<Line>{
    public int compare(Line o1, Line o2) {
        if(o1.getL() > o2.getL()){
            return 1;
        }else if(o1.getL() < o2.getL()){
            return -1;
        }else{
            return 0;
        }
    }
}

//线段
class Line {
    int l;
    int r;

    public Line(int l, int r){
        this.l = l;
        this.r = r;
    }

    public int getL() {
        return l;
    }

    public void setL(int l) {
        this.l = l;
    }

    public int getR() {
        return r;
    }

    public void setR(int r) {
        this.r = r;
    }

    public String toString(){
        return l + " " + r;
    }
}