Monocarp has got an array a consisting of n integers. Let's denote k as the mathematic mean of these elements (note that it's possible that k is not an integer). 

The mathematic mean of an array of n elements is the sum of elements divided by the number of these elements (i. e. sum divided by n).

Monocarp wants to delete exactly two elements from a so that the mathematic mean of the remaining (n - 2) elements is still equal to k.

Your task is to calculate the number of pairs of positions [i, j] (i < j) such that if the elements on these positions are deleted, the mathematic mean of (n - 2) remaining elements is equal to k (that is, it is equal to the mathematic mean of n elements of the original array a).

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.

The first line of each testcase contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.

The second line contains a sequence of integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{9}), where a_i is the i-th element of the array.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

Print one integer — the number of pairs of positions [i, j] (i < j) such that if the elements on these positions are deleted, the mathematic mean of (n - 2) remaining elements is equal to k (that is, it is equal to the mathematic mean of n elements of the original array a).

Example

Input


4
4
8 8 8 8
3
50 20 10
5
1 4 7 3 5
7
1 2 3 4 5 6 7


Output


6
0
2
3

Note

In the first example, any pair of elements can be removed since all of them are equal.

In the second example, there is no way to delete two elements so the mathematic mean doesn't change.

In the third example, it is possible to delete the elements on positions 1 and 3, or the elements on positions 4 and 5.
1
3
228 1337 228

0

from __future__ import division, print_function
import math
import sys
import os
from io import BytesIO, IOBase
#from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right

BUFSIZE = 8192


class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None

	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(b"\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()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
	"""Prints the values to a stream, or to sys.stdout by default."""
	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()


if sys.version_info[0] < 3:
	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

def I():
    return(int(input()))
def lint():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))
import itertools
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9+7)
import sys
# def I(): return int(stdin.readline())
# def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return list(map(str,input().strip()))
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter, deque
import math
import heapq
from heapq import heappop , heappush
import bisect
from math import factorial, inf
from itertools import groupby
from itertools import permutations as comb
def gcd(a,b): 
    while b:
        a %= b
        tmp = a
        a = b
        b = tmp
    
    return a
 
def lcm(a,b): 
    return a  // gcd(a, b) * b
 
def check_prime(n):
    for i in range(2, int(n ** (1 / 2)) + 1):
        if not n % i:
            return False
    return True

def nCr(n, r):
 
    return (fact(n) // (fact(r)
                * fact(n - r)))
 
# Returns factorial of n
def fact(n):
 
    res = 1
     
    for i in range(2, n+1):
        res = res * i
         
    return res
def primefactors(n):
    num=0
    
    while n % 2 == 0:
        num+=1
        n = n / 2
    
    for i in range(3,int(math.sqrt(n))+1,2):
         
    
        while n % i== 0:
            num+=1
            n = n // i
               
    
    if n > 2:
        num+=1
    return num
'''
def iter_ds(src):
    store=[src]
    while len(store):
        tmp=store.pop()
        if not vis[tmp]:
            vis[tmp]=True
            for j in ar[tmp]:
                store.append(j)
'''
def ask(a,b,c):
    # print('? 1 {}'.format(a),flush=True)
    print(c,a,b,flush=True)
    n=I()
    
    return n
def linear_sieve(n):
    is_composite=[False]*n
    prime=[]
    for i in range(2,n):
        if not is_composite[i]:
            prime.append(i)
        for j in prime:
            is_composite[i*j]=True
            if i%prime==0:
                break
    return prime
 
def dfs(i,p,d):
    
    a,tmp=0,0
    for j in d[i]:
        if j!=p:
            a+=1
            tmp+=dfs(j,i)
    
    if a==0:
        return 0
    
    return tmp/a + 1 
def primeFactors(n):
    l=[]
    
    while n % 2 == 0:
        l.append(2)
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2):
         
        
        while n % i== 0:
            l.append(i)
            n = n // i

    if n > 2:
        l.append(n)
    return l


    

#  Sieve
d=[]
primes=[]
prim=[0]*(10**5+1)
def sieve(n):
    
    for i in range(n):
        d.append(i)

    for i in range(2,n):

        if d[i]==i:
            prim[i]=1
            primes.append(i)
        j=0
        while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
        
            d[i * primes[j]] = primes[j]
            j+=1

        
def primeFactors(n):
    factors=[]
    while n!=1:
        factors.append(d[n])
        n//=d[n]
    
    return factors


mod=1000000007

t = 1 
t = I()
for _ in range(t):
    n = I()
    s = lint()
    ans=0
    k=sum(s)/n
    
    d = defaultdict(lambda:0)
    for i in range(n):
        d[s[i]]+=1
    for i in range(n):
        if (2*k-s[i])%1!=0:
            continue
        tmp=d[int(2*k-s[i])]
        if tmp>0:
            if s[i]==k:
                ans+=tmp-1
            else:
                ans+=tmp
    print(ans//2)
    # print((6*((n**(2**n -2))%mod))%mod)
    
    
       

            

        
