// for ceil x/y -- (x+y-1)/y
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define pll pair<ll,ll>
#define pi 3.14159265358979323846
const ll INF=1e18;
const ll MOD=1e9+7;
using namespace std;


ll POWER(ll base,ll expo)
{
    ll ret=1;
    while(expo)
    {
        if(expo&1)ret=ret*base%MOD;
        expo>>=1;
        base=base*base%MOD;
    }
    return ret%MOD;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll test_case=1; cin>>test_case;
    while(test_case--)
    {
        ll n; cin>>n;
        ll a[n];
        for(ll i=0;i<n;i++)cin>>a[i];

        vector<ll> cnt(62,0);
        for(ll i=0;i<n;i++){
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j))cnt[j]++;
            }
        }

        ll ans=0;

        for(ll i=0;i<n;i++){
            ll A=0;
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j)){
                    ll p=(1LL<<j)%MOD;
                    A=(A + cnt[j]*p)%MOD;
                }
            }
            ll O=0;
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j)){
                    ll p=(1LL<<j)%MOD;
                    O=(O + n*p)%MOD;
                }
                else{
                    ll p=(1LL<<j)%MOD;
                    O=(O + cnt[j]*p)%MOD;
                }
            }
            ans=(ans + A*O)%MOD;
        }
        cout<<ans<<endl;
    }
    return 0;
}

### 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
// for ceil x/y -- (x+y-1)/y
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define pll pair<ll,ll>
#define pi 3.14159265358979323846
const ll INF=1e18;
const ll MOD=1e9+7;
using namespace std;


ll POWER(ll base,ll expo)
{
    ll ret=1;
    while(expo)
    {
        if(expo&1)ret=ret*base%MOD;
        expo>>=1;
        base=base*base%MOD;
    }
    return ret%MOD;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll test_case=1; cin>>test_case;
    while(test_case--)
    {
        ll n; cin>>n;
        ll a[n];
        for(ll i=0;i<n;i++)cin>>a[i];

        vector<ll> cnt(62,0);
        for(ll i=0;i<n;i++){
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j))cnt[j]++;
            }
        }

        ll ans=0;

        for(ll i=0;i<n;i++){
            ll A=0;
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j)){
                    ll p=(1LL<<j)%MOD;
                    A=(A + cnt[j]*p)%MOD;
                }
            }
            ll O=0;
            for(ll j=0;j<62;j++){
                if(a[i]&(1LL<<j)){
                    ll p=(1LL<<j)%MOD;
                    O=(O + n*p)%MOD;
                }
                else{
                    ll p=(1LL<<j)%MOD;
                    O=(O + cnt[j]*p)%MOD;
                }
            }
            ans=(ans + A*O)%MOD;
        }
        cout<<ans<<endl;
    }
    return 0;
}

```