/// You just can't beat the person who never gives up
/// ICPC next year

#include<bits/stdc++.h>
using namespace std ;

const int N = 5e5+5 ,mod = 1e9+7 ;

int t ,n ;
long long a[N] ,ans_or[N] ,cnt[60] ;
vector<int> idx[60] ;
int main(){
    scanf("%d",&t);
    while(t--){
        for(int i=0;i<60;++i) idx[i].clear();
        memset(cnt,0,sizeof cnt);
        scanf("%d",&n);
        for(int i=0;i<n;++i){
            scanf("%lld",a+i);
            for(int j=59;j>=0;--j) if(a[i]>>j&1){
                idx[j].push_back(i);
                ++cnt[j] ;
            }
        }
        for(int i=0;i<n;++i){
            ans_or[i] = 0 ;
            for(int j=59;j>=0;--j){
                if(a[i]>>j&1){
                    long long cur = (1ll<<(1ll*j))%mod ;
                    cur = (1ll*cur*n)%mod ;
                    ans_or[i] = (cur+ans_or[i])%mod ;
                }
                else{
                    long long cur = (1ll<<(1ll*j))%mod ;
                    cur = (1ll*cur*cnt[j])%mod ;
                    ans_or[i] = (cur+ans_or[i])%mod ;
                }
            }
        }
        long long ans = 0 ;
        for(int j=59;j>=0;--j){
            for(int i=0;i<idx[j].size();++i){
                long long cur = (1ll<<(1ll*j))%mod ;
                cur = 1ll*cur*idx[j].size()%mod ;
                cur = 1ll*cur*ans_or[idx[j][i]]%mod;
                ans = (ans + cur)%mod ;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}

### Prompt

Generate 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
/// You just can't beat the person who never gives up
/// ICPC next year

#include<bits/stdc++.h>
using namespace std ;

const int N = 5e5+5 ,mod = 1e9+7 ;

int t ,n ;
long long a[N] ,ans_or[N] ,cnt[60] ;
vector<int> idx[60] ;
int main(){
    scanf("%d",&t);
    while(t--){
        for(int i=0;i<60;++i) idx[i].clear();
        memset(cnt,0,sizeof cnt);
        scanf("%d",&n);
        for(int i=0;i<n;++i){
            scanf("%lld",a+i);
            for(int j=59;j>=0;--j) if(a[i]>>j&1){
                idx[j].push_back(i);
                ++cnt[j] ;
            }
        }
        for(int i=0;i<n;++i){
            ans_or[i] = 0 ;
            for(int j=59;j>=0;--j){
                if(a[i]>>j&1){
                    long long cur = (1ll<<(1ll*j))%mod ;
                    cur = (1ll*cur*n)%mod ;
                    ans_or[i] = (cur+ans_or[i])%mod ;
                }
                else{
                    long long cur = (1ll<<(1ll*j))%mod ;
                    cur = (1ll*cur*cnt[j])%mod ;
                    ans_or[i] = (cur+ans_or[i])%mod ;
                }
            }
        }
        long long ans = 0 ;
        for(int j=59;j>=0;--j){
            for(int i=0;i<idx[j].size();++i){
                long long cur = (1ll<<(1ll*j))%mod ;
                cur = 1ll*cur*idx[j].size()%mod ;
                cur = 1ll*cur*ans_or[idx[j][i]]%mod;
                ans = (ans + cur)%mod ;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}

```