#include <bits/stdc++.h>
#define ll long long
#define pp push_back
#define ff first
#define ss second

using namespace std;

ll w,m,h,r,l,ii,jj,x,y[500001][61],z,zz[61],g,aa,bb,a,b,c,d,vis,k,mod=1e9+7,q,inf=1e18;
ll T,i,j,n;
vector <ll> v,vv;
pair <ll,ll> p;
map <ll,ll> mp,mpp,mppp;
set <pair <ll,ll> > se;
set <pair <ll,ll> > :: iterator it;
string s,t;
char cc;

int main()
{
    ios::sync_with_stdio(0);
    cin >> T;
    while(T--)
    {
        cin >> n;
        for(i=0;i<61;i++)
            zz[i]=0;
        for(i=0;i<n;i++)
        {
            cin >> x;
            a=x;
            j=0;
            while(a>0)
            {
                y[i][j]=a%2;
                zz[j]+=a%2;
                a/=2;
                j++;
            }
        }
        z=0;
        for(i=0;i<n;i++)
        {
            a=0;
            b=0;
            k=1;
            for(j=0;j<60;j++)
            {
                a+=((k)%mod*y[i][j]*zz[j])%mod;
                a%=mod;
                if(y[i][j]==1)
                    b+=((k%mod)*n)%mod;
                else
                    b+=((k%mod)*zz[j])%mod;
                b%=mod;
                k*=2;
                y[i][j]=0;
            }
            z+=(a*b)%mod;
            z%=mod;
        }
        cout << z;
        cout << endl;
    }
    return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:

Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i   \&   x_j) ⋅ (x_j   |   x_k)$$$

Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)

Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.

The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.

The sum of n over all test cases will not exceed 5 ⋅ 10^5.

Output

Print t lines. The i-th line should contain the answer to the i-th text case.

Example

Input


8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973


Output


128
91
1600
505
0
1
502811676
264880351

### Response

```cpp
#include <bits/stdc++.h>
#define ll long long
#define pp push_back
#define ff first
#define ss second

using namespace std;

ll w,m,h,r,l,ii,jj,x,y[500001][61],z,zz[61],g,aa,bb,a,b,c,d,vis,k,mod=1e9+7,q,inf=1e18;
ll T,i,j,n;
vector <ll> v,vv;
pair <ll,ll> p;
map <ll,ll> mp,mpp,mppp;
set <pair <ll,ll> > se;
set <pair <ll,ll> > :: iterator it;
string s,t;
char cc;

int main()
{
    ios::sync_with_stdio(0);
    cin >> T;
    while(T--)
    {
        cin >> n;
        for(i=0;i<61;i++)
            zz[i]=0;
        for(i=0;i<n;i++)
        {
            cin >> x;
            a=x;
            j=0;
            while(a>0)
            {
                y[i][j]=a%2;
                zz[j]+=a%2;
                a/=2;
                j++;
            }
        }
        z=0;
        for(i=0;i<n;i++)
        {
            a=0;
            b=0;
            k=1;
            for(j=0;j<60;j++)
            {
                a+=((k)%mod*y[i][j]*zz[j])%mod;
                a%=mod;
                if(y[i][j]==1)
                    b+=((k%mod)*n)%mod;
                else
                    b+=((k%mod)*zz[j])%mod;
                b%=mod;
                k*=2;
                y[i][j]=0;
            }
            z+=(a*b)%mod;
            z%=mod;
        }
        cout << z;
        cout << endl;
    }
    return 0;
}

```