#include<bits/stdc++.h>
using namespace std;
#define  int long long 
#define mod 1000000007
#define endl '\n'

int a[500005],cnt[64];

int mul(int a,int b) {
    return ((a%mod)*(b%mod))%mod;
}
int add(int a,int b) {
    return ((a%mod) + (b%mod))%mod;
}

void setBitAtIndex(int N) {
    memset(cnt,0,sizeof cnt);
    for(int i=0;i<N;i++) {
        for(int j=0;j<62;j++) {
            if((a[i]>>j)&1) {
                cnt[j]++;
            }
        }
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T=1;
    cin>>T;
    while(T--) {
        int N;
        cin>>N;
        int b[N],c[N];
        for(int i=0;i<N;i++) {
            cin>>a[i];
        }
        
        setBitAtIndex(N);
        
        // b is storing & and c is stroing | 
        memset(b,0,sizeof b);
        memset(c,0,sizeof c);
        int d[64];
        d[0] = 1;
        for(int i=1;i<64;i++) {
            d[i] = mul(d[i-1],(int)2);
        }
        for(int i=0;i<N;i++) {
            for(int j=62;j>=0;j--) {
                if((a[i]>>j)&1) {
                    b[i] = add(b[i],mul(d[j],cnt[j]));  
                    c[i] = add(c[i],mul(d[j],N));
                }
                else {
                    c[i] = add(c[i],mul(d[j],cnt[j]));
                }
            }
        }
        
        int ans = 0;
        for(int i=0;i<N;i++) {
            ans = add(ans,mul(b[i],c[i]));
        }
        cout<<ans<<endl;
    }
    return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
#define  int long long 
#define mod 1000000007
#define endl '\n'

int a[500005],cnt[64];

int mul(int a,int b) {
    return ((a%mod)*(b%mod))%mod;
}
int add(int a,int b) {
    return ((a%mod) + (b%mod))%mod;
}

void setBitAtIndex(int N) {
    memset(cnt,0,sizeof cnt);
    for(int i=0;i<N;i++) {
        for(int j=0;j<62;j++) {
            if((a[i]>>j)&1) {
                cnt[j]++;
            }
        }
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T=1;
    cin>>T;
    while(T--) {
        int N;
        cin>>N;
        int b[N],c[N];
        for(int i=0;i<N;i++) {
            cin>>a[i];
        }
        
        setBitAtIndex(N);
        
        // b is storing & and c is stroing | 
        memset(b,0,sizeof b);
        memset(c,0,sizeof c);
        int d[64];
        d[0] = 1;
        for(int i=1;i<64;i++) {
            d[i] = mul(d[i-1],(int)2);
        }
        for(int i=0;i<N;i++) {
            for(int j=62;j>=0;j--) {
                if((a[i]>>j)&1) {
                    b[i] = add(b[i],mul(d[j],cnt[j]));  
                    c[i] = add(c[i],mul(d[j],N));
                }
                else {
                    c[i] = add(c[i],mul(d[j],cnt[j]));
                }
            }
        }
        
        int ans = 0;
        for(int i=0;i<N;i++) {
            ans = add(ans,mul(b[i],c[i]));
        }
        cout<<ans<<endl;
    }
    return 0;
}
```