#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int NRBITS = 60;
const int MOD = 1e9 + 7;

int t, n;
ll x[500005];
ll are[NRBITS];
ll put[2 * NRBITS];

void getPut() {
    put[0] = 1;
    for (int i = 1; i < 2 * NRBITS; i++) {
        put[i] = (put[i - 1] << 1LL);
        if (put[i] >= MOD)
            put[i] -= MOD;
    }
}

int main()
{
    getPut();

    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%I64d", &x[i]);


        memset(are, 0, sizeof(are));
        for (int a = 0; a < NRBITS; a++)
            for (int i = 1; i <= n; i++)
                if (x[i] & (1LL << a))
                    are[a]++;

        ll ansFinal = 0;
        for (int j = 1; j <= n; j++) {
            ll sumAnd = 0;
            for (int bit = 0; bit < NRBITS; bit++) {
                if (!(x[j] & (1LL << bit)))
                    continue;
                sumAnd = (sumAnd + 1LL * put[bit] * are[bit] % MOD) % MOD;
            }

            ll sumOr = 0;
            for (int bit = 0; bit < NRBITS; bit++) {
                if (!(x[j] & (1LL << bit)))
                    sumOr = (sumOr + 1LL * put[bit] * are[bit] % MOD) % MOD;
                else
                    sumOr = (sumOr + 1LL * put[bit] * n % MOD) % MOD;
            }

            ansFinal = (ansFinal + sumAnd * sumOr % MOD) % MOD;
        }
        printf("%I64d\n", ansFinal);
    }

    return 0;
}

### 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
#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int NRBITS = 60;
const int MOD = 1e9 + 7;

int t, n;
ll x[500005];
ll are[NRBITS];
ll put[2 * NRBITS];

void getPut() {
    put[0] = 1;
    for (int i = 1; i < 2 * NRBITS; i++) {
        put[i] = (put[i - 1] << 1LL);
        if (put[i] >= MOD)
            put[i] -= MOD;
    }
}

int main()
{
    getPut();

    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%I64d", &x[i]);


        memset(are, 0, sizeof(are));
        for (int a = 0; a < NRBITS; a++)
            for (int i = 1; i <= n; i++)
                if (x[i] & (1LL << a))
                    are[a]++;

        ll ansFinal = 0;
        for (int j = 1; j <= n; j++) {
            ll sumAnd = 0;
            for (int bit = 0; bit < NRBITS; bit++) {
                if (!(x[j] & (1LL << bit)))
                    continue;
                sumAnd = (sumAnd + 1LL * put[bit] * are[bit] % MOD) % MOD;
            }

            ll sumOr = 0;
            for (int bit = 0; bit < NRBITS; bit++) {
                if (!(x[j] & (1LL << bit)))
                    sumOr = (sumOr + 1LL * put[bit] * are[bit] % MOD) % MOD;
                else
                    sumOr = (sumOr + 1LL * put[bit] * n % MOD) % MOD;
            }

            ansFinal = (ansFinal + sumAnd * sumOr % MOD) % MOD;
        }
        printf("%I64d\n", ansFinal);
    }

    return 0;
}

```