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.
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def readTree(n,e): # to read tree
    adj=[set() for i in range(n+1)]
    for i in range(e):
        u1,u2=IP()
        adj[u1].add(u2)
    return adj
def sieve():
    li=[True]*(10**3+5)
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime,cur=[0]*200,0
    for i in range(10**3+5):
        if li[i]==True:
            prime[cur]=i
            cur+=1
    return prime
def SPF():
    mx=(10**6+1)
    spf[1]=1
    for i in range(2,mx):
        if spf[i]==1e9:
            spf[i]=i
            for j in range(i*i,mx,i):
                if i<spf[j]:
                    spf[j]=i
    return
def prime(n,d):
    prm=set()
    while n!=1:
        prm.add(spf[n])
        n=n//spf[n]
    for ele in prm:
        d[ele]=d.get(ele,0)+1
    return
#####################################################################################
mod=998244353
def solve():
    n,k=IP()
    li=[]
    for i in range(n):
        a,b=IP()
        li.append([a,b])
    for i in range(n):
        found=True
        for j in range(n):
            if i!=j:
                dist=abs(li[i][0]-li[j][0])+abs(li[i][1]-li[j][1])
                if dist>k:
                    found=False
                    break
        if not(found):
            print(-1)
            return
    print(1)
    return


t=II()
for i in range(t):
    solve()

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