#include <cstdio>
#include <cstring>
using namespace std;

const int mod = 1e9 + 7;

#define int long long 

const int maxn = 5e5 + 10;
int a[maxn], cnt[63][2], n;

int read() {
  int x = 0; char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >='0' && ch <='9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}

int pow(const int &x, const int &y) {
  return (x % mod) * y % mod;
}

inline void update(int &x, const int& y) {
  x += y % mod; if (x >= mod) x -= mod;
}

inline void solve() {
  memset(cnt, 0, sizeof(cnt));
  n = read(); int ans = 0;
  for (int i = 1; i <= n; ++ i) a[i] = read();
  for (int i = 1; i <= n; ++ i) {
    for (int j = 0; j < 60; ++ j) {
      if(a[i] & (1ll << j)) cnt[j][1] ++;
      else cnt[j][0] ++;
    }
  }
  for (int i = 1; i <= n; ++ i) {
    int v1 = 0, v2 = 0;
    for (int j = 0; j < 60; ++ j) {
      if ((1ll << j) & a[i]) update(v1, pow((1ll << j), cnt[j][1]));
      update(v2, pow((1ll << j), cnt[j][1]));
      if ((1ll << j) & a[i]) update(v2, pow((1ll << j), cnt[j][0]));
    }
    update(ans, pow(v1, v2));
  }
  printf("%lld\n",ans);
}

signed main() {
  int t; t = read();
  while(t --) solve();
}

### Prompt

Please provide a CPP coded solution 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 <cstdio>
#include <cstring>
using namespace std;

const int mod = 1e9 + 7;

#define int long long 

const int maxn = 5e5 + 10;
int a[maxn], cnt[63][2], n;

int read() {
  int x = 0; char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >='0' && ch <='9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}

int pow(const int &x, const int &y) {
  return (x % mod) * y % mod;
}

inline void update(int &x, const int& y) {
  x += y % mod; if (x >= mod) x -= mod;
}

inline void solve() {
  memset(cnt, 0, sizeof(cnt));
  n = read(); int ans = 0;
  for (int i = 1; i <= n; ++ i) a[i] = read();
  for (int i = 1; i <= n; ++ i) {
    for (int j = 0; j < 60; ++ j) {
      if(a[i] & (1ll << j)) cnt[j][1] ++;
      else cnt[j][0] ++;
    }
  }
  for (int i = 1; i <= n; ++ i) {
    int v1 = 0, v2 = 0;
    for (int j = 0; j < 60; ++ j) {
      if ((1ll << j) & a[i]) update(v1, pow((1ll << j), cnt[j][1]));
      update(v2, pow((1ll << j), cnt[j][1]));
      if ((1ll << j) & a[i]) update(v2, pow((1ll << j), cnt[j][0]));
    }
    update(ans, pow(v1, v2));
  }
  printf("%lld\n",ans);
}

signed main() {
  int t; t = read();
  while(t --) solve();
}

```