#include<bits/stdc++.h>
#define ll long long 
#define ld long double 
#define MOD 1000000007
#define f first 
#define s second 
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
const ll inf=1e18+5;
const int maxN=1e5+1;
bool comparator(pair<pair<ll,int>,int> a,pair<pair<ll,int>,int> b){
    return a.f.f>=b.f.f;
}
int main(){
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    IOS
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<ll> a(n),cnt(60);
        ll sum=0ll;
        for(int i=0;i<n;i++){
            cin>>a[i];
            for(int j=0;j<60;j++){
                if((a[i])&(1ll<<j)){
                    cnt[j]++;
                }
            }
            sum=(sum+(a[i]%MOD))%MOD;
        }
        vector<ll> b(n);
        for(int i=0;i<n;i++){
            for(int j=0;j<60;j++){
                if((a[i])&(1ll<<j)){
                    b[i]=(b[i]+((1ll<<j)%MOD)*cnt[j])%MOD;
                }
            }
        }
        ll ans=0ll;
        for(int i=0;i<n;i++){
            ans=(ans+(b[i]*((sum+((a[i]%MOD)*n)%MOD-b[i])%MOD))%MOD)%MOD;
        }
        ans=(ans+MOD)%MOD;
        cout<<ans<<endl;
    }
}
### Prompt

Your task is to create 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 ld long double 
#define MOD 1000000007
#define f first 
#define s second 
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
const ll inf=1e18+5;
const int maxN=1e5+1;
bool comparator(pair<pair<ll,int>,int> a,pair<pair<ll,int>,int> b){
    return a.f.f>=b.f.f;
}
int main(){
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    IOS
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<ll> a(n),cnt(60);
        ll sum=0ll;
        for(int i=0;i<n;i++){
            cin>>a[i];
            for(int j=0;j<60;j++){
                if((a[i])&(1ll<<j)){
                    cnt[j]++;
                }
            }
            sum=(sum+(a[i]%MOD))%MOD;
        }
        vector<ll> b(n);
        for(int i=0;i<n;i++){
            for(int j=0;j<60;j++){
                if((a[i])&(1ll<<j)){
                    b[i]=(b[i]+((1ll<<j)%MOD)*cnt[j])%MOD;
                }
            }
        }
        ll ans=0ll;
        for(int i=0;i<n;i++){
            ans=(ans+(b[i]*((sum+((a[i]%MOD)*n)%MOD-b[i])%MOD))%MOD)%MOD;
        }
        ans=(ans+MOD)%MOD;
        cout<<ans<<endl;
    }
}
```