//in the name of GOD
#include <bits/stdc++.h>
#include <cmath>
using namespace std;
#define ll long long int
#define mpp make_pair
#define pii pair<ll,ll>
#define pb push_back

const int MD=1e9+7;
ll pow2[70];

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    pow2[0]=1;
    for(int i=1;i<65;i++){
        pow2[i]=(pow2[i-1]*2)%MD;
        pow2[i]=(pow2[i]+MD)%MD;
    }
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        ll a[n+1];
        for(int i=0;i<n;i++) cin>>a[i];
        int bit[70];
        fill(bit,bit+70,0);
        for(int i=0;i<n;i++){
            ll k=a[i];
            int j=0;
            while(k>0){
                if(k%2==1) bit[j]++;
                j++;
                k/=2;
            }
        }
        ll ans=0;
        for(int i=0;i<n;i++){
            bool b[70];
            fill(b,b+70,0);
            ll k=a[i];
            int j=0;
            while(k>0){
                if(k%2==1) b[j]=1;
                j++;
                k/=2;
            }
            ll andi=0;
            for(int i=0;i<60;i++){
                if(b[i]==1){
                    andi+=(bit[i]*pow2[i])%MD;
                    andi=(andi+MD)%MD;
                }
            }
            ll ori=0;
            for(int i=0;i<60;i++){
                if(b[i]==1){
                    ori+=(n*pow2[i])%MD;
                    ori=(ori+MD)%MD;
                }
                else{
                    ori+=(bit[i]*pow2[i]%MD);
                    ori=(ori+MD)%MD;
                }
            }
            ll ppp=(andi*ori)%MD;
            ppp=(ppp+MD)%MD;
            ans=(ans+ppp)%MD;
            ans=(ans+MD)%MD;
        }
        cout<<ans<<endl;
    }
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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
//in the name of GOD
#include <bits/stdc++.h>
#include <cmath>
using namespace std;
#define ll long long int
#define mpp make_pair
#define pii pair<ll,ll>
#define pb push_back

const int MD=1e9+7;
ll pow2[70];

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    pow2[0]=1;
    for(int i=1;i<65;i++){
        pow2[i]=(pow2[i-1]*2)%MD;
        pow2[i]=(pow2[i]+MD)%MD;
    }
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        ll a[n+1];
        for(int i=0;i<n;i++) cin>>a[i];
        int bit[70];
        fill(bit,bit+70,0);
        for(int i=0;i<n;i++){
            ll k=a[i];
            int j=0;
            while(k>0){
                if(k%2==1) bit[j]++;
                j++;
                k/=2;
            }
        }
        ll ans=0;
        for(int i=0;i<n;i++){
            bool b[70];
            fill(b,b+70,0);
            ll k=a[i];
            int j=0;
            while(k>0){
                if(k%2==1) b[j]=1;
                j++;
                k/=2;
            }
            ll andi=0;
            for(int i=0;i<60;i++){
                if(b[i]==1){
                    andi+=(bit[i]*pow2[i])%MD;
                    andi=(andi+MD)%MD;
                }
            }
            ll ori=0;
            for(int i=0;i<60;i++){
                if(b[i]==1){
                    ori+=(n*pow2[i])%MD;
                    ori=(ori+MD)%MD;
                }
                else{
                    ori+=(bit[i]*pow2[i]%MD);
                    ori=(ori+MD)%MD;
                }
            }
            ll ppp=(andi*ori)%MD;
            ppp=(ppp+MD)%MD;
            ans=(ans+ppp)%MD;
            ans=(ans+MD)%MD;
        }
        cout<<ans<<endl;
    }
    return 0;
}

```