#include <bits/stdc++.h>

using namespace std;

#ifndef NJUDGE 
  #define dbg(...)
#endif

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
 
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ll NLINF = 0xf7f7f7f7f7f7f7f7;
const int INF = 0x3f3f3f3f, NINF = 0xf7f7f7f7; 
const int MOD1 = 1e9+7, MOD2 = 1e9+9;
 
#define sz(x) (int) x.size()
#define eb emplace_back
#define pb push_back
#define ss second
#define ff first

const int N = 5e5+10, B = 65;
ll a[N], orv[N], andv[N];
int t, n, freq[B][2];

int ith(ll v, int i){
  return (int) min((1ll << i) & v, 1ll);
}

void wipe(){
  for(int i = 1; i <= n; i++)
    orv[i] = andv[i] = 0ll;
  for(int j = 0; j <= 60; j++)
    freq[j][0] = freq[j][1] = 0;
}

int main(){
  for(scanf("%d", &t); t--;){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
      scanf("%lld", &a[i]);
      for(int j = 0; j <= 60; j++)
        freq[j][ith(a[i], j)]++;
    }

    for(int i = 1; i <= n; i++){
      for(int j = 0; j <= 60; j++){
        if (ith(a[i], j) == 1){
          andv[i] = (andv[i] + ((1ll << j) % MOD1)*freq[j][1] % MOD1) % MOD1;
          orv[i] = (orv[i] + ((1ll << j) % MOD1)*n % MOD1) % MOD1;
        } 
        else 
          orv[i] = (orv[i] + ((1ll << j) % MOD1)*freq[j][1] % MOD1) % MOD1;
      }
    }

    ll ans = 0ll;
    for(int i = 1; i <= n; i++)
      ans = (ans + 1ll*orv[i]*andv[i] % MOD1) % MOD1;

    printf("%lld\n", ans);
    wipe();
  }
}
### 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;

#ifndef NJUDGE 
  #define dbg(...)
#endif

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
 
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ll NLINF = 0xf7f7f7f7f7f7f7f7;
const int INF = 0x3f3f3f3f, NINF = 0xf7f7f7f7; 
const int MOD1 = 1e9+7, MOD2 = 1e9+9;
 
#define sz(x) (int) x.size()
#define eb emplace_back
#define pb push_back
#define ss second
#define ff first

const int N = 5e5+10, B = 65;
ll a[N], orv[N], andv[N];
int t, n, freq[B][2];

int ith(ll v, int i){
  return (int) min((1ll << i) & v, 1ll);
}

void wipe(){
  for(int i = 1; i <= n; i++)
    orv[i] = andv[i] = 0ll;
  for(int j = 0; j <= 60; j++)
    freq[j][0] = freq[j][1] = 0;
}

int main(){
  for(scanf("%d", &t); t--;){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
      scanf("%lld", &a[i]);
      for(int j = 0; j <= 60; j++)
        freq[j][ith(a[i], j)]++;
    }

    for(int i = 1; i <= n; i++){
      for(int j = 0; j <= 60; j++){
        if (ith(a[i], j) == 1){
          andv[i] = (andv[i] + ((1ll << j) % MOD1)*freq[j][1] % MOD1) % MOD1;
          orv[i] = (orv[i] + ((1ll << j) % MOD1)*n % MOD1) % MOD1;
        } 
        else 
          orv[i] = (orv[i] + ((1ll << j) % MOD1)*freq[j][1] % MOD1) % MOD1;
      }
    }

    ll ans = 0ll;
    for(int i = 1; i <= n; i++)
      ans = (ans + 1ll*orv[i]*andv[i] % MOD1) % MOD1;

    printf("%lld\n", ans);
    wipe();
  }
}
```