// https://codeforces.com/contest/1466/problem/E

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

typedef long long ll;

const ll M = 1e9 + 7;

#define IS(x, i) (((x) & (1ll << (i))) != 0)

ll add(ll a, ll b) { return ((a % M) + (b % M)) % M; }
ll mul(ll a, ll b) { return ((a % M) * (b % M)) % M; }

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int B = 60;
        int n;
        cin >> n;
        vector<ll> v(n);
        vector<int> nrset(64, 0);
        for (ll &x : v) {
            cin >> x;
            if (x > 0) {
                // int b = __builtin_clz(x);
                // B = max(B, 63 - b);
            }
            for (int i = 0; i < B; ++i) {
                if (IS(x, i))
                    nrset[i]++;
            }
        }
        ll ans = 0;
        for (int j = 0; j < n; ++j) {
            const ll xj = v[j];

            ll X = 0;
            for (int i = 0; i < B; ++i) {
                ll w = (1ll << i) % M;
                if (IS(xj, i))
                    X = add(X, mul(w, nrset[i]));
            }

            ll Y = 0;
            for (int i = 0; i < B; ++i) {
                int k = IS(xj, i) ? n : nrset[i];
                Y = add(Y, mul(k, 1ll << i));
            }

            ans = add(ans, mul(X, Y));
        }
        cout << ans << endl;
    }
    return 0;
}

### Prompt

Generate 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
// https://codeforces.com/contest/1466/problem/E

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

typedef long long ll;

const ll M = 1e9 + 7;

#define IS(x, i) (((x) & (1ll << (i))) != 0)

ll add(ll a, ll b) { return ((a % M) + (b % M)) % M; }
ll mul(ll a, ll b) { return ((a % M) * (b % M)) % M; }

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int B = 60;
        int n;
        cin >> n;
        vector<ll> v(n);
        vector<int> nrset(64, 0);
        for (ll &x : v) {
            cin >> x;
            if (x > 0) {
                // int b = __builtin_clz(x);
                // B = max(B, 63 - b);
            }
            for (int i = 0; i < B; ++i) {
                if (IS(x, i))
                    nrset[i]++;
            }
        }
        ll ans = 0;
        for (int j = 0; j < n; ++j) {
            const ll xj = v[j];

            ll X = 0;
            for (int i = 0; i < B; ++i) {
                ll w = (1ll << i) % M;
                if (IS(xj, i))
                    X = add(X, mul(w, nrset[i]));
            }

            ll Y = 0;
            for (int i = 0; i < B; ++i) {
                int k = IS(xj, i) ? n : nrset[i];
                Y = add(Y, mul(k, 1ll << i));
            }

            ans = add(ans, mul(X, Y));
        }
        cout << ans << endl;
    }
    return 0;
}

```