#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define INF 1000000000000000000
#define llu unsigned long long int
#define pb push_back
#define ff first
#define ss second
long long mod=1000000007;
/*
bool comparator(const pair<vector<int>,int> a,const pair<vector<int>,int> b)
{
return a.ff.size()<b.ff.size();
}
*/
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
lli t,i,j,n,ans,s,s1;
lli pow[60];
pow[0]=1;
for(i=1;i<=59;i++)
pow[i]=(pow[i-1]*2)%mod;
cin>>t;
while(t--)
{
  cin>>n;
  lli a[n];
  vector <lli> cnt(60,0);
  for(i=0;i<n;i++)
  {
  cin>>a[i];
  bitset<60> b(a[i]);
  for(j=0;j<60;j++)
  cnt[j]+=b[j];
  }
  ans=0;
  for(j=0;j<n;j++)
  {
    s=0;s1=0;
    bitset<60> b(a[j]);
    for(i=0;i<60;i++)
    {
    if(b[i])
    s=(s+pow[i]%mod*cnt[i])%mod;
    }
    for(i=0;i<60;i++)
    {
    if(b[i])
    s1=(s1+pow[i]%mod*n)%mod;
    else
    s1=(s1+pow[i]%mod*cnt[i])%mod;
    }
    ans=(ans+(s1*s)%mod)%mod;
  }
  cout<<ans<<endl;
}
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
#define lli long long int
#define INF 1000000000000000000
#define llu unsigned long long int
#define pb push_back
#define ff first
#define ss second
long long mod=1000000007;
/*
bool comparator(const pair<vector<int>,int> a,const pair<vector<int>,int> b)
{
return a.ff.size()<b.ff.size();
}
*/
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
lli t,i,j,n,ans,s,s1;
lli pow[60];
pow[0]=1;
for(i=1;i<=59;i++)
pow[i]=(pow[i-1]*2)%mod;
cin>>t;
while(t--)
{
  cin>>n;
  lli a[n];
  vector <lli> cnt(60,0);
  for(i=0;i<n;i++)
  {
  cin>>a[i];
  bitset<60> b(a[i]);
  for(j=0;j<60;j++)
  cnt[j]+=b[j];
  }
  ans=0;
  for(j=0;j<n;j++)
  {
    s=0;s1=0;
    bitset<60> b(a[j]);
    for(i=0;i<60;i++)
    {
    if(b[i])
    s=(s+pow[i]%mod*cnt[i])%mod;
    }
    for(i=0;i<60;i++)
    {
    if(b[i])
    s1=(s1+pow[i]%mod*n)%mod;
    else
    s1=(s1+pow[i]%mod*cnt[i])%mod;
    }
    ans=(ans+(s1*s)%mod)%mod;
  }
  cout<<ans<<endl;
}
}
```