Problem: Polycarp has invited n friends to celebrate the New Year. During the celebration, he decided to take a group photo of all his friends. Each friend can stand or lie on the side.

Each friend is characterized by two values h_i (their height) and w_i (their width). On the photo the i-th friend will occupy a rectangle h_i × w_i (if they are standing) or w_i × h_i (if they are lying on the side).

The j-th friend can be placed in front of the i-th friend on the photo if his rectangle is lower and narrower than the rectangle of the i-th friend. Formally, at least one of the following conditions must be fulfilled:

  * h_j < h_i and w_j < w_i (both friends are standing or both are lying); 
  * w_j < h_i and h_j < w_i (one of the friends is standing and the other is lying). 



For example, if n = 3, h=[3,5,3] and w=[4,4,3], then:

  * the first friend can be placed in front of the second: w_1 < h_2 and h_1 < w_2 (one of the them is standing and the other one is lying); 
  * the third friend can be placed in front of the second: h_3 < h_2 and w_3 < w_2 (both friends are standing or both are lying). 



In other cases, the person in the foreground will overlap the person in the background.

Help Polycarp for each i find any j, such that the j-th friend can be located in front of the i-th friend (i.e. at least one of the conditions above is fulfilled).

Please note that you do not need to find the arrangement of all people for a group photo. You just need to find for each friend i any other friend j who can be located in front of him. Think about it as you need to solve n separate independent subproblems.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of friends.

This is followed by n lines, each of which contains a description of the corresponding friend. Each friend is described by two integers h_i and w_i (1 ≤ h_i, w_i ≤ 10^9) — height and width of the i-th friend, respectively.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case output n integers on a separate line, where the i-th number is the index of a friend that can be placed in front of the i-th. If there is no such friend, then output -1.

If there are several answers, output any.

Example

Input


4
3
3 4
5 4
3 3
3
1 3
2 2
3 1
4
2 2
3 1
6 3
5 4
4
2 2
2 3
1 1
4 4


Output


-1 3 -1 
-1 -1 -1 
-1 -1 2 2 
3 3 -1 3 

Note

The first test case is described in the statement.

In the third test case, the following answers are also correct: 

  * [-1, -1, 1, 2]; 
  * [-1, -1, 1, 1]; 
  * [-1, -1, 2, 1]. 

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int max(int a, int b) {
    if (a > b) {
        return a;
    }
    else {
        return b;
    }
}

int Min(int a, int b) {
    if (a < b) {
        return a;
    }
    else {
        return b;
    }
}

vector <vector <int> > dims;
vector <int> res;

int main() {
    int t;
    cin >> t;
    int n;
    int temp1, temp2;
    int min, j, prevmin;
    while (t--) {
        cin >> n;
        res.resize(n);
        dims.resize(n);
        for (int i = 0; i < n; i++) {
            cin >> temp1 >> temp2;
            dims[i].push_back(i);
            dims[i].push_back(Min(temp1, temp2));
            dims[i].push_back(max(temp1, temp2));
        }

        sort(dims.begin(), dims.end(), [](const vector <int>& a, const vector <int>& b) {
            if (a[1] != b[1])
            {
                return a[1] < b[1];

            }
            else {
                return a[2] < b[2];
            }});

        j = 0;
        min = prevmin = 0;
        while (dims[j][1] == dims[0][1]) {
            res[dims[j][0]] = -1;
            j++;
            if (j == n) {
                break;
            }
        }

        if (j < n) {
            for (int i = j; i < n; i++) {
                if (dims[i][1] != dims[i - 1][1]) {
                    if (dims[min][2] > dims[prevmin][2]) {
                        min = prevmin;
                    }
                    prevmin = i;
                }

                if (dims[i][2] > dims[min][2]) {
                    res[dims[i][0]] = dims[min][0] + 1;
                }
                else {
                    res[dims[i][0]] = -1;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            cout << res[i] << " ";
        }

        cout << endl;

        dims.clear();
        res.clear();
    }
    return 0;
}