#include <bits/stdc++.h>

using namespace std;
#define LL long long
#define PB push_back
#define double long double
const double eps=1e-10,MAXN=1e18;
const int N=6e5+10,mod=1e9+7;
int n,q;
LL a[N];
LL s[N];
void solve(){
    scanf("%d",&n);
    for(int i=0;i<61;i++)s[i]=0;
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        for(int j=0;j<61;j++)if((a[i]>>j)&1)s[j]++;
    }
    LL ans=0;
    for(int i=1;i<=n;i++){
        //cout<<i<<endl;
        LL sum=0;
        for(int j=0;j<61;j++){
            if((a[i]>>j)&1)sum+=(1ll<<j)%mod*n%mod;
        else
            sum+=(1ll<<j)%mod*s[j]%mod;
            sum%=mod;
        }
    //cout<<i<<endl;
        for(int j=0;j<61;j++){
            if((a[i]>>j)&1){
                ans+=(1ll<<j)%mod*s[j]%mod*sum%mod;
                //cout<<i<<' '<<j<<' '<<ans<<' '<<sum<<' '<<(1ll<<j)<<' '<<s[j]<<endl;
            }
            ans%=mod;

        }
        //cout<<i<<' '<<ans<<endl;
    }
    printf("%lld\n",ans);
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
/*
2
0 1
1 1
1 0
0 1

3
1 0 0
0 1 0
0 0 1
1 1 1
1 1 1
1 1 1

*/

### Prompt

In CPP, your task is to solve 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 LL long long
#define PB push_back
#define double long double
const double eps=1e-10,MAXN=1e18;
const int N=6e5+10,mod=1e9+7;
int n,q;
LL a[N];
LL s[N];
void solve(){
    scanf("%d",&n);
    for(int i=0;i<61;i++)s[i]=0;
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        for(int j=0;j<61;j++)if((a[i]>>j)&1)s[j]++;
    }
    LL ans=0;
    for(int i=1;i<=n;i++){
        //cout<<i<<endl;
        LL sum=0;
        for(int j=0;j<61;j++){
            if((a[i]>>j)&1)sum+=(1ll<<j)%mod*n%mod;
        else
            sum+=(1ll<<j)%mod*s[j]%mod;
            sum%=mod;
        }
    //cout<<i<<endl;
        for(int j=0;j<61;j++){
            if((a[i]>>j)&1){
                ans+=(1ll<<j)%mod*s[j]%mod*sum%mod;
                //cout<<i<<' '<<j<<' '<<ans<<' '<<sum<<' '<<(1ll<<j)<<' '<<s[j]<<endl;
            }
            ans%=mod;

        }
        //cout<<i<<' '<<ans<<endl;
    }
    printf("%lld\n",ans);
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
/*
2
0 1
1 1
1 0
0 1

3
1 0 0
0 1 0
0 0 1
1 1 1
1 1 1
1 1 1

*/

```