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
/*#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("unroll-loops")*/
#include<bits/stdc++.h>
using namespace std;

#define  CIN            ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define  ll             long long int
#define  pb             push_back
#define  ull            unsigned long long int
#define  pii            pair < ll, ll>
#define  pll            pair < ll, pii>
#define  ff             first
#define  ss             second
#define  mx4            1000005
#define  MOD            1000000007
#define  mod2           1000000009
#define  min3(a,b,c)    min(a,min(b,c))
#define  min4(a,b,c,d)  min(d,min(a,min(b,c)))
#define  max3(a,b,c)    max(a,max(b,c))
#define  max4(a,b,c,d)  max(d,max(a,max(b,c)))
#define  ms(a,b)        memset(a,b,sizeof(a))
#define  mpp            make_pair
#define  gcd(a, b)      __gcd(a,b)
#define  lcm(a, b)      ((a)*(b)/gcd(a,b))
#define  input          freopen("input.txt","rt", stdin)
#define  output         freopen("output.txt","wt", stdout)
#define  PI             2*acos(0.0)
#define  zero(a)        memset(a,0,sizeof a)
#define  all(a)         a.begin(),a.end()
#define  rall(a)        a.rbegin(),a.rend()
#define  rev(a)         reverse(all(a))
#define  upper(s)       transform(s.begin(),s.end(),s.begin(), ::toupper)
#define  lower(s)       transform(s.begin(),s.end(),s.begin(), ::tolower)
#define  Unique(X)      (X).erase(unique(all(X)),(X).end())
#define  MX             200005
#define  nl             '\n'
#define  yes            cout<<"YES"<<endl
#define  no             cout<<"NO"<<endl
#define  fix(x)         cout<<fixed<<setprecision(x)
#define  coutv(v)       for(auto it:v)cout<<it<<space;cout<<nl;
#define  cinv(v)        for(auto &it:v)cin>>it;
#define  Case(x)        cout<<"Case "<<x<<": "
#define  strtoint(a)    atoi(a.c_str())
#define  dbg(x)         cerr<<#x<<" is "<<x<<'\n';
#define  fix(x)         cout<<fixed<<setprecision(x)
#define  leastonepos(x) __builtin_ffs(x)
#define  leadingoff(x)  __builtin_clz(x)
#define  trailingoff(x) __builtin_ctz(x)
#define  numofone(x)    __builtin_popcount(x)
#define  error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

///Compute_phi_co_prime
/*void computeTotient(int n){ll phi[n+5];for (int i=1; i<=n; i++)phi[i] = i;for (int p=2; p<=n; p++)
{if (phi[p] == p){phi[p] = p-1;for (int i = 2*p; i<=n; i += p){phi[i] = (phi[i]/p) * (p-1);}}}}*/
///seieve
/*vector<bool>mark(100000004);vector<int>prm;void sieve(int x){mark[0]=mark[1]=1;
for(int i =4; i<=x; i=i+2)mark[i]=1;for(int i=3; i*i<=x; i+=2)if(!mark[i])
for(int j=i*i; j<=x; j+=2*i)mark[j]=1;prm.push_back(2);for(int i=3; i<=x; i+=2)
if(mark[i]==0)prm.push_back(i);}*/
///PrimeFactorization
/*vector<int>pf;void primeFact(ll n){for(int i=0; i<prm.size()&&prm[i]*prm[i]<=n; i++)
{if(n%prm[i]==0){while(n%prm[i]==0){n=n/prm[i] ;pf.push_back(prm[i]);}}}if(n>1)pf.push_back(n);}*/
///power
/*ll pow( ll base , ll val ){ll res = 1 ;for(int i = 1 ; i <= val ; i++ ){res *= base ;}return res ;}*/
///Fibonacci
/*ll fib[ 100005 ] ;fib[ 1 ] = 1 ;fib[ 2 ] = 2 ;
for( int i = 3 ; i <= 100000 ; i ++ ){ fib[ i ] = ( fib[ i - 1 ] + fib[ i - 2 ] ) % MOD ;}*/
//min_dis(character)
/*ll min_dis(char ch1,char ch2){ll x = abs(ch1-ch2);return min(x,26-x);}*/
///leap_year
/*inline bool isLeapYear(ll year) { return (year%400==0) || (year%4==0 && year%100!=0); }*/
///BinToDec
/*ll BinToDec(bitset<64>bs)
{ll sum = 0 ,lim=bs.size();for( int i = 0 ; i <= lim ; i ++ )sum += bs2[ i ] * pow( 2, i ) ;return sum ;}*/
///Multiplication_With_Mod
/*long long fun(long long a, long long b){long long res = 0;a %= MOD;while (b)
{if (b & 1)res = (res + a) % MOD;a = (2 * a) % MOD;b >>= 1;}return res;}*/
///power of two
/*ll Pow[65];void power_of_two(){Pow[0]=1;
for(int i = 1 ; i <= 60 ; i++) Pow[i]=Pow[i-1]*2ll;}*/
///---------------Order set-------------------
///*os.find_by_order(index) os.order_of_key(value)
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using orderset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
orderset<ll>os ;*/
/*ll Os_Bound(ll x , ll UpOrLow)
{if(UpOrLow==2) x++;ll pos = os.order_of_key(x);if (pos == os.size()) return -1;else return *(os.find_by_order(pos));}*/

///priority_queue<int,vector<int>,greater<int> >pq;
///string s5 = "abcdefghijklmnopqrstuvwxyz";
///string s6 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
///int days [14] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
///string month[14]= {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};
///--------------Graph Moves--------------------------------------
///const int fx[] = {+1,-1,+0,+0};
///const int fy[] = {+0,+0,+1,-1};
///const int fx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's move
///const int fy[] = {-1,+1,+0,+0,+1,+1,-1,-1}; ///king's Move
///const int fx[] = {-2,-2,-1,-1,+1,+1,+2,+2}; ///knight's move
///const int fy[] = {-1,+1,-2,+2,-2,+2,-1,+1}; ///knight's move
///---------------------------------------------------------------
void err(istream_iterator<string> it)
{
    cerr<<endl;
}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
{
    cerr << *it << " = " << a << " | ";
    err(++it, args...);
}
/*int Set(int N,int pos){return N=N | (1<<pos);}
bool check(int N,int pos){return (bool)(N & (1<<pos));}
int reset(int N,int pos){return N= N & ~(1<<pos);}*/
/*
/*ll bigmod(ll a , ll b , ll m)
{
    if(b==0) return 1;
    if(b&1) return (a%m * bigmod(a,b-1,m))%m;
    ll x = bigmod(a,b/2,m);
    return (x*x)%m;
}
ll inverse(ll a , ll b , ll m)
{
    return ((a%m)*bigmod(b,m-2,m)%m);
}*/
/*
set<ll>st;
set<ll>::reverse_itertor it ;
for( it = rbegin() ; it!=st.rend() ; it++)
st.erase(next(it).base());
set<ll>::iterator it ;
for( it = begin() ; it!=st.end() ; it++)
st.erase(it);
*/
/*
struct data{
    ll wt,iq,indx;
    bool operator <(const data &other) const{
        if (wt==other.wt) return iq>other.iq;
        return wt<other.wt;
    }
}ara[1005];
*/

vector<ll>v1,v2,v3,v4;
///vector<pii>v5;
string s1, s2, s3;
///map<ll,ll>mp;
///map<ll,ll>::iterator it;
///ll ara[100005];
///multiset<ll>ms;
///multiset<ll>::iterator it;
///set<ll>st;
///set<ll>::iterator it;

int main()
{
    CIN;
    ll  a,b,c,d,e,f,i,j,k,m,n,cnt1=0,cnt2=0,sum1=0,sum2=0,chk1=0,chk2=0,T=1;
    cin>> T ;
    while(T--)
    {
        v1.clear();
        cin >> a ;
        for( i = 0 ; i < a ; i++)
        {
            cin >> b ;
            v1.pb(b);
        }
        sort(all(v1));
        ll ans = 0 ;
        for( i = 0 ; i < a ; i++)
        {
            ll x = upper_bound(all(v1),v1[i]+2)-v1.begin();
            if(x-i>=2)
            {
                ll ex = x-i;
                ex-=2;
                ex=(ex*(ex+1))/2;
                ans+=ex;

            }
        }
        cout<<ans<<endl;

    }
    return 0 ;
}
///printf & cout ekshatey use korbo nah
