#include <math.h>
#include <stdio.h>

#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

#define int ll
#define PI acos(-1.0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pii;
typedef vector<pii> vpii;

const ll MOD = 1000000007;

ll t, n;
ll x[1000000];
ll da[100];
ll tmp;

signed main() {
    cout.precision(15);
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>t;
    while (t--) {
        cin>>n;
        for (int i = 0; i < 80; i++) da[i] = 0;
        for (int i = 0; i < n; i++) {
            cin>>tmp;
            for (int j = 0; j < 60; j++) {
                if (tmp & (1LL<<j)) {
                    da[j] += (1LL<<j);
                    da[j] %= MOD;
                }
            }
            x[i] = tmp;
        }
        int ans = 0;
        for (int j = 0; j < n; j++) {
            int sum1 = 0;
            int sum2 = 0;
            tmp = x[j];
            for (int i = 0; i < 60; i++) {
                if (tmp & (1LL << i)) {
                    sum1 += da[i];
                    sum2 += ((1LL<<i)%MOD) * n;
                    sum2 %= MOD;
                    sum1 %= MOD;
                } else {
                    sum2 += da[i];
                    sum2 %= MOD;
                }
            }
            sum1 %= MOD;
            sum2 %= MOD;
            ans += sum1 * sum2;
            ans %= MOD;
        }
        cout << ans << endl;
    }
    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 <math.h>
#include <stdio.h>

#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

#define int ll
#define PI acos(-1.0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pii;
typedef vector<pii> vpii;

const ll MOD = 1000000007;

ll t, n;
ll x[1000000];
ll da[100];
ll tmp;

signed main() {
    cout.precision(15);
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>t;
    while (t--) {
        cin>>n;
        for (int i = 0; i < 80; i++) da[i] = 0;
        for (int i = 0; i < n; i++) {
            cin>>tmp;
            for (int j = 0; j < 60; j++) {
                if (tmp & (1LL<<j)) {
                    da[j] += (1LL<<j);
                    da[j] %= MOD;
                }
            }
            x[i] = tmp;
        }
        int ans = 0;
        for (int j = 0; j < n; j++) {
            int sum1 = 0;
            int sum2 = 0;
            tmp = x[j];
            for (int i = 0; i < 60; i++) {
                if (tmp & (1LL << i)) {
                    sum1 += da[i];
                    sum2 += ((1LL<<i)%MOD) * n;
                    sum2 %= MOD;
                    sum1 %= MOD;
                } else {
                    sum2 += da[i];
                    sum2 %= MOD;
                }
            }
            sum1 %= MOD;
            sum2 %= MOD;
            ans += sum1 * sum2;
            ans %= MOD;
        }
        cout << ans << endl;
    }
    return 0;
}

```