#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;
long long v[500005],fv1[62],fv0[62],p[62],maxim;

void calc(long long x)
{
    long long i=0,y=maxim;
    while(y!=0)
    {
        fv1[i]+=x%2;
        fv0[i]+=1-(x%2);
        x/=2;
        y/=2;
        i++;
    }
}

int main()
{
    long long t,l,n,i,sol,s,k,s1,s2,x,y;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    p[0]=1;
    for(i=1;i<=60;i++)
        p[i]=p[i-1]*2,p[i]%=MOD;;
    cin>>t;
    for(l=1;l<=t;l++)
    {
        cin>>n;
        s=0;
        sol=0;
        maxim=0;
        for(i=1;i<=n;i++)
            cin>>v[i],s+=v[i],s%=MOD,maxim=max(maxim,v[i]);
        for(i=1;i<=n;i++)
            calc(v[i]);
        for(i=1;i<=n;i++)
        {
            x=v[i];
            k=0;
            s1=s;
            s2=s;
            y=maxim;
            while(y!=0)
            {
                if(x%2==1)
                {
                    s2+=(fv0[k]*p[k])%MOD;
                    s2%=MOD;
                }
                else
                {
                    s1-=(fv1[k]*p[k])%MOD;
                    s1%=MOD;
                    if(s1<0)
                        s1+=MOD;
                }
                y/=2;
                x/=2;
                k++;
            }
            sol+=(s1*s2)%MOD;
            sol%=MOD;
        }
        cout<<sol<<'\n';
        for(i=0;i<=60;i++)
            fv0[i]=fv1[i]=0;
    }
    return 0;
}

### Prompt

Generate 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 MOD 1000000007
using namespace std;
long long v[500005],fv1[62],fv0[62],p[62],maxim;

void calc(long long x)
{
    long long i=0,y=maxim;
    while(y!=0)
    {
        fv1[i]+=x%2;
        fv0[i]+=1-(x%2);
        x/=2;
        y/=2;
        i++;
    }
}

int main()
{
    long long t,l,n,i,sol,s,k,s1,s2,x,y;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    p[0]=1;
    for(i=1;i<=60;i++)
        p[i]=p[i-1]*2,p[i]%=MOD;;
    cin>>t;
    for(l=1;l<=t;l++)
    {
        cin>>n;
        s=0;
        sol=0;
        maxim=0;
        for(i=1;i<=n;i++)
            cin>>v[i],s+=v[i],s%=MOD,maxim=max(maxim,v[i]);
        for(i=1;i<=n;i++)
            calc(v[i]);
        for(i=1;i<=n;i++)
        {
            x=v[i];
            k=0;
            s1=s;
            s2=s;
            y=maxim;
            while(y!=0)
            {
                if(x%2==1)
                {
                    s2+=(fv0[k]*p[k])%MOD;
                    s2%=MOD;
                }
                else
                {
                    s1-=(fv1[k]*p[k])%MOD;
                    s1%=MOD;
                    if(s1<0)
                        s1+=MOD;
                }
                y/=2;
                x/=2;
                k++;
            }
            sol+=(s1*s2)%MOD;
            sol%=MOD;
        }
        cout<<sol<<'\n';
        for(i=0;i<=60;i++)
            fv0[i]=fv1[i]=0;
    }
    return 0;
}

```