#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <math.h>
#include <deque>
#include <iomanip>
#include <fstream>
#define ll long long
#define ld long double

using namespace std;

const ll mod = 1e9 + 7;
const int lg = 60;

ll add(ll a, ll b) {
    a += b;
    while (a >= mod)
        a -= mod;
    return a;
}

void solve() {
    int n;
    cin >> n;
    vector<ll> x(n);
    vector<ll> cnt(lg);
    for (int i = 0; i < n; ++i) {
        cin >> x[i];
        for (int j = 0; j < lg; ++j) {
            if (x[i] & ((ll)1 << j)) {
                cnt[j]++;
            }
        }
    }
    ll ans = 0;
    for (int j = 0; j < n; ++j) {
        ll s1 = 0;
        for (int i = 0; i < lg; ++i) {
            ll val = (ll)1 << i;
            if (x[j] & val) {
                s1 = add(s1, (val % mod) * cnt[i] % mod);
            }
        }
        ll s2 = 0;
        for (int i = 0; i < lg; ++i) {
            ll val = (ll)1 << i;
            if (x[j] & val) {
                s2 = add(s2, (val % mod) * n % mod);
            } else {
                s2 = add(s2, (val % mod) * cnt[i] % mod);
            }
        }
        s1 *= s2;
        s1 %= mod;
        ans = add(ans, s1);
    }
    cout << ans << "\n";
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    for (int ij = 0; ij < t; ++ij) {
        solve();
    }
}
//*/
### 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
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <math.h>
#include <deque>
#include <iomanip>
#include <fstream>
#define ll long long
#define ld long double

using namespace std;

const ll mod = 1e9 + 7;
const int lg = 60;

ll add(ll a, ll b) {
    a += b;
    while (a >= mod)
        a -= mod;
    return a;
}

void solve() {
    int n;
    cin >> n;
    vector<ll> x(n);
    vector<ll> cnt(lg);
    for (int i = 0; i < n; ++i) {
        cin >> x[i];
        for (int j = 0; j < lg; ++j) {
            if (x[i] & ((ll)1 << j)) {
                cnt[j]++;
            }
        }
    }
    ll ans = 0;
    for (int j = 0; j < n; ++j) {
        ll s1 = 0;
        for (int i = 0; i < lg; ++i) {
            ll val = (ll)1 << i;
            if (x[j] & val) {
                s1 = add(s1, (val % mod) * cnt[i] % mod);
            }
        }
        ll s2 = 0;
        for (int i = 0; i < lg; ++i) {
            ll val = (ll)1 << i;
            if (x[j] & val) {
                s2 = add(s2, (val % mod) * n % mod);
            } else {
                s2 = add(s2, (val % mod) * cnt[i] % mod);
            }
        }
        s1 *= s2;
        s1 %= mod;
        ans = add(ans, s1);
    }
    cout << ans << "\n";
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    for (int ij = 0; ij < t; ++ij) {
        solve();
    }
}
//*/
```