You have n distinct points (x_1, y_1),…,(x_n,y_n) on the plane and a non-negative integer parameter k. Each point is a microscopic steel ball and k is the attract power of a ball when it's charged. The attract power is the same for all balls.

In one operation, you can select a ball i to charge it. Once charged, all balls with Manhattan distance at most k from ball i move to the position of ball i. Many balls may have the same coordinate after an operation.

More formally, for all balls j such that |x_i - x_j| + |y_i - y_j| ≤ k, we assign x_j:=x_i and y_j:=y_i.

<image> An example of an operation. After charging the ball in the center, two other balls move to its position. On the right side, the red dot in the center is the common position of those balls. 

Your task is to find the minimum number of operations to move all balls to the same position, or report that this is impossible.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains two integers n, k (2 ≤ n ≤ 100, 0 ≤ k ≤ 10^6) — the number of balls and the attract power of all balls, respectively.

The following n lines describe the balls' coordinates. The i-th of these lines contains two integers x_i, y_i (0 ≤ x_i, y_i ≤ 10^5) — the coordinates of the i-th ball.

It is guaranteed that all points are distinct.

Output

For each test case print a single integer — the minimum number of operations to move all balls to the same position, or -1 if it is impossible.

Example

Input


3
3 2
0 0
3 3
1 1
3 3
6 7
8 8
6 9
4 1
0 0
0 1
0 2
0 3


Output


-1
1
-1

Note

In the first test case, there are three balls at (0, 0), (3, 3), and (1, 1) and the attract power is 2. It is possible to move two balls together with one operation, but not all three balls together with any number of operations.

In the second test case, there are three balls at (6, 7), (8, 8), and (6, 9) and the attract power is 3. If we charge any ball, the other two will move to the same position, so we only require one operation.

In the third test case, there are four balls at (0, 0), (0, 1), (0, 2), and (0, 3), and the attract power is 1. We can show that it is impossible to move all balls to the same position with a sequence of operations.
#!/usr/bin/env pypy
from __future__ import division, print_function
from collections import defaultdict, Counter, deque
from future_builtins import ascii, filter, hex, map, oct, zip
from itertools import imap as map, izip as zip, permutations, combinations, combinations_with_replacement
from __builtin__ import xrange as range
from math import ceil, factorial, log, sqrt
from _continuation import continulet
from cStringIO import StringIO
from io import IOBase
import __pypy__
from bisect import bisect, insort, bisect_left, bisect_right
from fractions import Fraction
import heapq
from functools import reduce
import string
import sys
import os
import re
inf = float('inf')
mod = int(1e9) + 7
mod_ = 998244353

def dist(a, b):
    return abs(a[0] - b[0]) + abs(a[1] - b[1])

def solve():
    n, k = map(int, input().split())
    arr = [tuple(map(int, input().split())) for i in range(n)]
    steps = 0
    i = 0
    while i < len(arr):
        changed = False
        for j in range(len(arr)):
            if dist(arr[i], arr[j]) <= k:
                arr[j] = arr[i]
                changed = True
        if changed:
            steps += 1
            arr = list(set(arr))
        i += 1
    if len(arr) == 1:
        print(steps)
    else:
        print(-1)



def main():
    for _ in range(int(input())):
        solve()

BUFSIZE = 8192
class FastI(IOBase):
    def __init__(self, file):
        self._fd = file.fileno()
        self._buffer = StringIO()
        self.newlines = 0

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count("\n") + (not b)
            ptr = self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(
                b), self._buffer.seek(ptr)
        self.newlines -= 1
        return self._buffer.readline()
class FastO(IOBase):
    def __init__(self, file):
        self._fd = file.fileno()
        self._buffer = __pypy__.builders.StringBuilder()
        self.write = lambda s: self._buffer.append(s)

    def flush(self):
        os.write(self._fd, self._buffer.build())
        self._buffer = __pypy__.builders.StringBuilder()
def print(*args, **kwargs):
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
sys.stdin, sys.stdout = FastI(sys.stdin), FastO(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
    def bootstrap(cont):
        call, arg = cont.switch()
        while True:
            call, arg = cont.switch(to=continulet(
                lambda _, f, args: f(*args), call, arg))
    cont = continulet(bootstrap)
    cont.switch()
    main()