#include <iostream>
#include <algorithm>

constexpr int N = 500005;
constexpr int p = 1000000007;

inline int add(int x, int y) { return (x += y) >= p ? x - p : x; }
inline int mul(int x, int y) { return (long long)x * y % p; }

int T, n;
long long a[N];
int cnt[60];

int main() {
    std::ios::sync_with_stdio(false), std::cin.tie(nullptr);
    std::cin >> T;
    while (T--) {
        std::cin >> n;
        std::fill(cnt, cnt + 60, 0);
        for (int i = 1; i <= n; ++i) {
            std::cin >> a[i];
            for (int j = 0; j < 60; ++j) if (a[i] >> j & 1) ++cnt[j];
        }
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int x = 0, y = 0;
            for (int j = 0, pow = 1; j < 60; ++j, pow = mul(pow, 2))
                if ((a[i] >> j) & 1) x = add(x, mul(cnt[j], pow)), y = add(y, mul(n, pow));
                else y = add(y, mul(cnt[j], pow));
            ans = add(ans, mul(x, y));
        }
        std::cout << ans << '\n';
    }
    return 0;
}

### Prompt

Please formulate a cpp solution to 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 <iostream>
#include <algorithm>

constexpr int N = 500005;
constexpr int p = 1000000007;

inline int add(int x, int y) { return (x += y) >= p ? x - p : x; }
inline int mul(int x, int y) { return (long long)x * y % p; }

int T, n;
long long a[N];
int cnt[60];

int main() {
    std::ios::sync_with_stdio(false), std::cin.tie(nullptr);
    std::cin >> T;
    while (T--) {
        std::cin >> n;
        std::fill(cnt, cnt + 60, 0);
        for (int i = 1; i <= n; ++i) {
            std::cin >> a[i];
            for (int j = 0; j < 60; ++j) if (a[i] >> j & 1) ++cnt[j];
        }
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int x = 0, y = 0;
            for (int j = 0, pow = 1; j < 60; ++j, pow = mul(pow, 2))
                if ((a[i] >> j) & 1) x = add(x, mul(cnt[j], pow)), y = add(y, mul(n, pow));
                else y = add(y, mul(cnt[j], pow));
            ans = add(ans, mul(x, y));
        }
        std::cout << ans << '\n';
    }
    return 0;
}

```