#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll MOD = 1e9+7;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
void add(ll &a, ll b) {
    a = (a + b) % MOD;
}
int main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<ll> a(n);
        rep(i, n) cin >> a[i];
        vector<ll> count(60, 0);
        rep(i, n) {
            rep(j, 60) {
                if(a[i] >> j & 1) {
                    count[j]++;
                }
            }
        }
        ll ans = 0;
        vector<ll> two(60, 0);
        two[0] = 1;
        vector<ll> twoc(60, 0);
        for(int i = 1; i < 60; i++) two[i] = (two[i-1] * 2) % MOD;
        rep(i, 60) twoc[i] = two[i] * count[i] % MOD;
        rep(i, n) {
            ll x = 0;
            ll y = 0;
            rep(j, 60) {
                if(a[i] >> j & 1) {
                    add(x, twoc[j]);
                    add(y, two[j] * n % MOD);
                } else {
                    add(y, twoc[j]);
                }
            }
            add(ans, x * y % MOD);
        }
        cout << ans << endl;
    }
}

### 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 <bits/stdc++.h>
using namespace std;
using ll = long long;
ll MOD = 1e9+7;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
void add(ll &a, ll b) {
    a = (a + b) % MOD;
}
int main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<ll> a(n);
        rep(i, n) cin >> a[i];
        vector<ll> count(60, 0);
        rep(i, n) {
            rep(j, 60) {
                if(a[i] >> j & 1) {
                    count[j]++;
                }
            }
        }
        ll ans = 0;
        vector<ll> two(60, 0);
        two[0] = 1;
        vector<ll> twoc(60, 0);
        for(int i = 1; i < 60; i++) two[i] = (two[i-1] * 2) % MOD;
        rep(i, 60) twoc[i] = two[i] * count[i] % MOD;
        rep(i, n) {
            ll x = 0;
            ll y = 0;
            rep(j, 60) {
                if(a[i] >> j & 1) {
                    add(x, twoc[j]);
                    add(y, two[j] * n % MOD);
                } else {
                    add(y, twoc[j]);
                }
            }
            add(ans, x * y % MOD);
        }
        cout << ans << endl;
    }
}

```