/*
  author : Anand
*/

#include <bits/stdc++.h>

using namespace std;

#define PI 3.1415926535897932384626
#define int long long
#define ll long long

#ifndef ONLINE_JUDGE
#define debug(...) cerr << "\t" << #__VA_ARGS__ << " : " << (__VA_ARGS__) << endl;
#else
#define debug(...) 42
#endif

const int INF = 1e18 + 5;
const int MOD = 1000000007LL;
const int N = 1e7;
const int K = 25;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<int> bitCnt(60);
    for (int i = 0; i < 60; ++i)
        for (int x : a)
            if ((1LL << i) & x) ++bitCnt[i];
    vector<int> twoPowUnderMod(60);
    twoPowUnderMod[0] = 1;
    for (int i = 1; i < 60; ++i) twoPowUnderMod[i] = (2 * twoPowUnderMod[i - 1]) % MOD;
    int sumUnderMod = 0;
    for (int j : a) sumUnderMod = (sumUnderMod + j) % MOD;
    int ans = 0;
    for (int j : a) {
        int sumAnds = 0, sumOrs = 0;
        for (int i = 0; i < 60; ++i)
            if ((1LL << i) & j)
                sumAnds = (sumAnds + (bitCnt[i] * twoPowUnderMod[i]) % MOD) % MOD;
        sumOrs = sumUnderMod;
        sumOrs = (sumOrs + (j % MOD) * (n % MOD)) % MOD;
        sumOrs = (sumOrs - sumAnds + MOD) % MOD;
        int toAdd = ((sumAnds % MOD) * (sumOrs % MOD)) % MOD;
        ans += toAdd;
        ans %= MOD;
    }
    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }

    return 0;
}
### Prompt

In Cpp, your task is to solve 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
/*
  author : Anand
*/

#include <bits/stdc++.h>

using namespace std;

#define PI 3.1415926535897932384626
#define int long long
#define ll long long

#ifndef ONLINE_JUDGE
#define debug(...) cerr << "\t" << #__VA_ARGS__ << " : " << (__VA_ARGS__) << endl;
#else
#define debug(...) 42
#endif

const int INF = 1e18 + 5;
const int MOD = 1000000007LL;
const int N = 1e7;
const int K = 25;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<int> bitCnt(60);
    for (int i = 0; i < 60; ++i)
        for (int x : a)
            if ((1LL << i) & x) ++bitCnt[i];
    vector<int> twoPowUnderMod(60);
    twoPowUnderMod[0] = 1;
    for (int i = 1; i < 60; ++i) twoPowUnderMod[i] = (2 * twoPowUnderMod[i - 1]) % MOD;
    int sumUnderMod = 0;
    for (int j : a) sumUnderMod = (sumUnderMod + j) % MOD;
    int ans = 0;
    for (int j : a) {
        int sumAnds = 0, sumOrs = 0;
        for (int i = 0; i < 60; ++i)
            if ((1LL << i) & j)
                sumAnds = (sumAnds + (bitCnt[i] * twoPowUnderMod[i]) % MOD) % MOD;
        sumOrs = sumUnderMod;
        sumOrs = (sumOrs + (j % MOD) * (n % MOD)) % MOD;
        sumOrs = (sumOrs - sumAnds + MOD) % MOD;
        int toAdd = ((sumAnds % MOD) * (sumOrs % MOD)) % MOD;
        ans += toAdd;
        ans %= MOD;
    }
    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }

    return 0;
}
```