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
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 FIP(): # to take tuple as input
    return map(float,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
def primeFactors(n,d):

    # Print the number of two's that divide n
    while n % 2 == 0:
        d[2]=d.get(2,0)+1
        n = n / 2

    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3,int(mt.sqrt(n))+1,2):

        # while i divides n , print i ad divide n
        while n % i== 0:
            d[i]=d.get(i,0)+1
            n = n / i

    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        d[n]=d.get(n,0)+1
    return
#####################################################################################
mod=998244353
def solve():
    n=II()
    a=L()
    ans=0
    d={}
    for ele in a:
        d[ele]=d.get(ele,0)+1
    a.sort()
    #print(d)
    #print(a)
    for ele in a:
        last=ans
        ans+=(d.get(ele+1,0)*d.get(ele+2,0))
        if d.get(ele+1,0)>=2:
            val=d[ele+1]
            ans+=((val*(val-1))//2)
        if d.get(ele+2,0)>=2:
            val=d[ele+2]
            ans+=((val*(val-1))//2)
        if d.get(ele,0)>=2:
            val=d[ele]
            ans+=((val-1)*d.get(ele+1,0))
        if d.get(ele,0)>=2:
            val=d[ele]
            ans+=((val-1)*d.get(ele+2,0))
        if d.get(ele,0)>=3:
            val=d[ele]
            ans+=(((val-2)*(val-1))//2)
        d[ele]-=1
       # print("here ",ans-last)
    print(ans)
    return

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

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