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

const int N = 500000 + 9;
const long long M = 1000000007;

long long a[N];
long long cnt[100], b[100];
long long Pow2[100];


void solve(){
    int n; scanf("%d", &n);
    memset(cnt, 0, sizeof(cnt));
    memset(b, 0, sizeof(b));
    for (int i = 1; i <= n; ++i){
        scanf("%lld", a + i);
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1) ++cnt[p];
        }
    }
    for (int i = 1; i <= n; ++i){
        long long sum = 0;
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1){
                sum = (sum + n * Pow2[p] % M) % M;
            }
            else {
                sum = (sum + cnt[p] * Pow2[p] % M) % M;
            }
        }
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1){
                b[p] = (b[p] + sum) % M;
            }
        }
    }
    long long ans = 0;
    for (int p = 0; p < 60; ++p){
        ans = (ans + Pow2[p] * cnt[p] % M * b[p] % M) % M;
    }
    printf("%lld\n", ans);
}

int main(){
    Pow2[0] = 1;
    for (int i = 1; i <= 60; ++i){
        Pow2[i] = Pow2[i - 1] * 2 % M;
    }
    int T; scanf("%d", &T);
    while (T--) solve();
    return 0;
}
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;

const int N = 500000 + 9;
const long long M = 1000000007;

long long a[N];
long long cnt[100], b[100];
long long Pow2[100];


void solve(){
    int n; scanf("%d", &n);
    memset(cnt, 0, sizeof(cnt));
    memset(b, 0, sizeof(b));
    for (int i = 1; i <= n; ++i){
        scanf("%lld", a + i);
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1) ++cnt[p];
        }
    }
    for (int i = 1; i <= n; ++i){
        long long sum = 0;
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1){
                sum = (sum + n * Pow2[p] % M) % M;
            }
            else {
                sum = (sum + cnt[p] * Pow2[p] % M) % M;
            }
        }
        for (int p = 0; p < 60; ++p){
            if (a[i] >> p & 1){
                b[p] = (b[p] + sum) % M;
            }
        }
    }
    long long ans = 0;
    for (int p = 0; p < 60; ++p){
        ans = (ans + Pow2[p] * cnt[p] % M * b[p] % M) % M;
    }
    printf("%lld\n", ans);
}

int main(){
    Pow2[0] = 1;
    for (int i = 1; i <= 60; ++i){
        Pow2[i] = Pow2[i - 1] * 2 % M;
    }
    int T; scanf("%d", &T);
    while (T--) solve();
    return 0;
}
```