#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define all(a) (a).begin(), (a).end()
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define fr first
#define sc second
#define int long long
#define rc(s) return cout<<s,0
#define rcc(s) cout<<s,exit(0)

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int nmax = 500005;

const int mod = 1e9 + 7;

int t, n, a[nmax], ans, put[60], rasp[nmax];

vector <int> cnt[60];

int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
    //freopen("input.in", "r", stdin);
    put[0] = 1;
    for(int i=1;i<=59;i++) put[i] = 1LL * 2 * put[i - 1] % mod;
    cin >> t;
    while(t--){
        ans = 0;
        cin >> n;
        for(int i=1;i<=n;i++){
            cin >> a[i];
        }
        for(int i=0;i<60;i++) cnt[i].clear();
        for(int i=1;i<=n;i++) rasp[i] = 0;
        for(int i=0;i<60;i++){
            for(int j=1;j<=n;j++){
                if(a[j] & (1LL << i)){
                    cnt[i].push_back(j);
                }
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<60;j++){
                if(a[i] & (1LL << j)){
                    rasp[i] = (rasp[i] + n * put[j] % mod) % mod;
                }
                else{
                    rasp[i] = (rasp[i] + (int)cnt[j].size() * put[j] % mod) % mod;
                }
            }
        }
        for(int i=0;i<=60;i++){
            int curr = 0;
            for(auto it : cnt[i]){
                curr = rasp[it];
                curr = curr * (put[i] % mod) % mod;
                curr = curr * (int)cnt[i].size() % mod;
                ans = (ans + curr) % mod;
            }
        }
        cout << ans << '\n';
    }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
#define ll long long
#define ld long double
#define all(a) (a).begin(), (a).end()
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define fr first
#define sc second
#define int long long
#define rc(s) return cout<<s,0
#define rcc(s) cout<<s,exit(0)

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int nmax = 500005;

const int mod = 1e9 + 7;

int t, n, a[nmax], ans, put[60], rasp[nmax];

vector <int> cnt[60];

int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
    //freopen("input.in", "r", stdin);
    put[0] = 1;
    for(int i=1;i<=59;i++) put[i] = 1LL * 2 * put[i - 1] % mod;
    cin >> t;
    while(t--){
        ans = 0;
        cin >> n;
        for(int i=1;i<=n;i++){
            cin >> a[i];
        }
        for(int i=0;i<60;i++) cnt[i].clear();
        for(int i=1;i<=n;i++) rasp[i] = 0;
        for(int i=0;i<60;i++){
            for(int j=1;j<=n;j++){
                if(a[j] & (1LL << i)){
                    cnt[i].push_back(j);
                }
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<60;j++){
                if(a[i] & (1LL << j)){
                    rasp[i] = (rasp[i] + n * put[j] % mod) % mod;
                }
                else{
                    rasp[i] = (rasp[i] + (int)cnt[j].size() * put[j] % mod) % mod;
                }
            }
        }
        for(int i=0;i<=60;i++){
            int curr = 0;
            for(auto it : cnt[i]){
                curr = rasp[it];
                curr = curr * (put[i] % mod) % mod;
                curr = curr * (int)cnt[i].size() % mod;
                ans = (ans + curr) % mod;
            }
        }
        cout << ans << '\n';
    }
}

```