#include<bits/stdc++.h>
using namespace std;
// cout << fixed << setprecision(12) << p;
#define ll long long
#define ld long double
#define pb push_back

#define ins insert
#define mp make_pair
#define ff first
#define ss second
#define pi  pair<int,int>
#define pl  pair<ll,ll>
#define all(v)  v.begin(),v.end()
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define mod 1000000007
const int N = 100000 + 5;

ll modexp(ll x, ll n, ll m = mod)
{
    if (n == 0) {
        return 1;
    }
    else if (n % 2 == 0) {
        return modexp((x * x) % m, n / 2);
    }
    else {
        return (x * modexp((x * x) % m, (n - 1) / 2) % m);
    }
}


ll add(ll a, ll b) {
    return ((a % mod) + (b % mod)) % mod;
}
ll mul(ll a, ll b) {
    return ((a % mod) * (b % mod)) % mod;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fio;

    ll t;
    cin >> t;

    while (t--)
    {
        ll n, i, j, x, y, ans = 0, p;
        cin >> n;

        ll a[n];

        for (i = 0; i < n; i++)
            cin >> a[i];

        ll bit[65] = {0};

        for (i = 0; i < n; i++)
        {
            p = 1;
            for (j = 1; j <= 60; j++)
            {
                if (a[i] & p)
                    bit[j]++;
                p = p * 2;
            }
        }


        for (i = 0; i < n; i++)
        {
            x = 0, y = 0;
            p = 1;
            for (j = 1; j <= 60; j++)
            {

                if (a[i] & p)
                {
                    x = add( x , mul((bit[j]) , p ) );

                }
                if (a[i] & p)
                {
                    y = add( y , mul(n, p )  );

                }
                else
                {
                    y = add( y , mul((bit[j]) , p )  );
                }

                // cout << x << " " << y << endl;

                p = p * 2;

            }

            ans = add(ans, mul(x , y)) ;
        }

        cout << ans << endl;




    }


}


### Prompt

In cpp, your task is to solve 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>
using namespace std;
// cout << fixed << setprecision(12) << p;
#define ll long long
#define ld long double
#define pb push_back

#define ins insert
#define mp make_pair
#define ff first
#define ss second
#define pi  pair<int,int>
#define pl  pair<ll,ll>
#define all(v)  v.begin(),v.end()
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define mod 1000000007
const int N = 100000 + 5;

ll modexp(ll x, ll n, ll m = mod)
{
    if (n == 0) {
        return 1;
    }
    else if (n % 2 == 0) {
        return modexp((x * x) % m, n / 2);
    }
    else {
        return (x * modexp((x * x) % m, (n - 1) / 2) % m);
    }
}


ll add(ll a, ll b) {
    return ((a % mod) + (b % mod)) % mod;
}
ll mul(ll a, ll b) {
    return ((a % mod) * (b % mod)) % mod;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fio;

    ll t;
    cin >> t;

    while (t--)
    {
        ll n, i, j, x, y, ans = 0, p;
        cin >> n;

        ll a[n];

        for (i = 0; i < n; i++)
            cin >> a[i];

        ll bit[65] = {0};

        for (i = 0; i < n; i++)
        {
            p = 1;
            for (j = 1; j <= 60; j++)
            {
                if (a[i] & p)
                    bit[j]++;
                p = p * 2;
            }
        }


        for (i = 0; i < n; i++)
        {
            x = 0, y = 0;
            p = 1;
            for (j = 1; j <= 60; j++)
            {

                if (a[i] & p)
                {
                    x = add( x , mul((bit[j]) , p ) );

                }
                if (a[i] & p)
                {
                    y = add( y , mul(n, p )  );

                }
                else
                {
                    y = add( y , mul((bit[j]) , p )  );
                }

                // cout << x << " " << y << endl;

                p = p * 2;

            }

            ans = add(ans, mul(x , y)) ;
        }

        cout << ans << endl;




    }


}


```