#include <bits/stdc++.h>

#define pb push_back
#define eb emplace_back
#define F first
#define S second

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())

using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;

template<class T> bool ckmin(T& a, T b) { return b < a ? a = b, true : false; }
template<class T> bool ckmax(T& a, T b) { return b > a ? a = b, true : false; }

const int MOD = 1e9+7;

void solve() {
    int n; cin >> n;
    vector<ll> v(n);
    vi cnt(60);
    for (int j = n; j--; ) {
        ll x; cin >> x; v[j] = x;
        for (int i = 0; x != 0; ++i) {
            if (x % 2) ++cnt[i];
            x /= 2;
        }
    }

    int ans = 0;
    for (int j = n; j--; ) {
        int x = 0;
        for (int i = 0, pi = 1; i < 60; ++i) {
            if ((v[j] >> i) & 1) x = (x + (ll)pi * cnt[i]) % MOD;
            pi = 2 * pi % MOD;
        }
        for (int i = 0, pi = 1; i < 60; ++i) {
            int c = ((v[j] >> i) & 1) ? n : cnt[i];
            ans = ((ll)pi*c%MOD*x + ans) % MOD;
            pi = 2 * pi % MOD;
        }
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int t = 1; cin >> t;
    while(t--) solve();
}
### Prompt

Your task is to create 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>

#define pb push_back
#define eb emplace_back
#define F first
#define S second

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())

using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;

template<class T> bool ckmin(T& a, T b) { return b < a ? a = b, true : false; }
template<class T> bool ckmax(T& a, T b) { return b > a ? a = b, true : false; }

const int MOD = 1e9+7;

void solve() {
    int n; cin >> n;
    vector<ll> v(n);
    vi cnt(60);
    for (int j = n; j--; ) {
        ll x; cin >> x; v[j] = x;
        for (int i = 0; x != 0; ++i) {
            if (x % 2) ++cnt[i];
            x /= 2;
        }
    }

    int ans = 0;
    for (int j = n; j--; ) {
        int x = 0;
        for (int i = 0, pi = 1; i < 60; ++i) {
            if ((v[j] >> i) & 1) x = (x + (ll)pi * cnt[i]) % MOD;
            pi = 2 * pi % MOD;
        }
        for (int i = 0, pi = 1; i < 60; ++i) {
            int c = ((v[j] >> i) & 1) ? n : cnt[i];
            ans = ((ll)pi*c%MOD*x + ans) % MOD;
            pi = 2 * pi % MOD;
        }
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int t = 1; cin >> t;
    while(t--) solve();
}
```