#include <bits/stdc++.h>
#define ll long long
#define ull int64_t
#define vll vector<ll>
using namespace std;
ull mod = 1e9 + 7;
ull g(int c){
  return (((ull)(1)) << c);
}
int f(ull x, int c) {
  if (x & g(c))
    return 1;
  return 0;
}
inline void sol() {
  int n;
  cin >> n;
  vector<ull> x(n);
  for (auto &i : x)
    cin >> i;
  vector<int> fre(60);
  for (int j = 0; j < 60; j++) {
    for (int i = 0; i < n; i++) {
      if (f(x[i], j))
        fre[j]++;
    }
  }
  ull ans = 0;
  for (int j = 0; j < n; j++) {
    ull p1 = 0, p2 = 0;
    for (ull c = 0; c < 60; c++) {
      p1 += (g(c) % mod) * f(x[j], c) * fre[c];
      p1 %= mod;
      p2 += (g(c) % mod) * ((n - (1 - f(x[j], c)) * (n - fre[c])) % mod);
      p2 %= mod;
    }
    ans += ((p1 % mod) * (p2 % mod)) % mod;
    ans %= mod;
  }
  cout << ans << endl;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int t;
  cin >> t;
  while (t--)
    sol();

  return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 ll long long
#define ull int64_t
#define vll vector<ll>
using namespace std;
ull mod = 1e9 + 7;
ull g(int c){
  return (((ull)(1)) << c);
}
int f(ull x, int c) {
  if (x & g(c))
    return 1;
  return 0;
}
inline void sol() {
  int n;
  cin >> n;
  vector<ull> x(n);
  for (auto &i : x)
    cin >> i;
  vector<int> fre(60);
  for (int j = 0; j < 60; j++) {
    for (int i = 0; i < n; i++) {
      if (f(x[i], j))
        fre[j]++;
    }
  }
  ull ans = 0;
  for (int j = 0; j < n; j++) {
    ull p1 = 0, p2 = 0;
    for (ull c = 0; c < 60; c++) {
      p1 += (g(c) % mod) * f(x[j], c) * fre[c];
      p1 %= mod;
      p2 += (g(c) % mod) * ((n - (1 - f(x[j], c)) * (n - fre[c])) % mod);
      p2 %= mod;
    }
    ans += ((p1 % mod) * (p2 % mod)) % mod;
    ans %= mod;
  }
  cout << ans << endl;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int t;
  cin >> t;
  while (t--)
    sol();

  return 0;
}
```