This is the easy version of this problem. The only difference between easy and hard versions is the constraints on k and m (in this version k=2 and m=3). Also, in this version of the problem, you DON'T NEED to output the answer by modulo.

You are given a sequence a of length n consisting of integers from 1 to n. The sequence may contain duplicates (i.e. some elements can be equal).

Find the number of tuples of m = 3 elements such that the maximum number in the tuple differs from the minimum by no more than k = 2. Formally, you need to find the number of triples of indices i < j < z such that

$$$max(a_i, a_j, a_z) - min(a_i, a_j, a_z) ≤ 2.$$$

For example, if n=4 and a=[1,2,4,3], then there are two such triples (i=1, j=2, z=4 and i=2, j=3, z=4). If n=4 and a=[1,1,1,1], then all four possible triples are suitable.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases. Then t test cases follow.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the sequence a.

The next line contains n integers a_1, a_2,…, a_n (1 ≤ a_i ≤ n) — the sequence a.

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

Output

Output t answers to the given test cases. Each answer is the required number of triples of elements, such that the maximum value in the triple differs from the minimum by no more than 2. Note that in difference to the hard version of the problem, you don't need to output the answer by modulo. You must output the exact value of the answer.

Example

Input


4
4
1 2 4 3
4
1 1 1 1
1
1
10
5 6 1 3 2 9 8 1 2 4


Output


2
4
0
15
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 

"""**************************MAIN*****************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        a.sort()
        d=[0]*n
        ans=0
        for i in a:
            d[i-1]+=1
            x=d[i-1]
            if i>1:
                x+=d[i-2]
            if i>2:
                x+=d[i-3]
            x=((x-1)*(x-2))//2
            ans+=max(0,x)
        print(ans)




"""***********************************************"""

######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()