// Sometimes, the questions are complicated - and the answers are simple. //
#include<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld  long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int N=5e5+9, mod=1e9+7;;
int t, n, cnt[61], val[61];
ll a[N];
int main() {
    IO

    for(ll j=0 ; j<60 ; j++)
        val[j]=(1LL<<j)%mod;

    cin>>t;
    while(t--) {
        cin>>n;

        ll ans=0, s=0;
        memset(cnt,0,sizeof cnt);

        for(int i=1; i<=n; i++) {
            cin>>a[i];
            s=(s+a[i]%mod);
            for(ll j=0 ; j<60 ; j++)
                if(((1LL<<j)&a[i])!=0)
                    cnt[j]++;
        }

        s%=mod;

        for(int i=1; i<=n; i++) {
            ll sor=0;
            ll sum2=0;
            for(int j=0; j<60; j++) {
                if(((1LL<<j)&a[i])!=0) {

                    ans=(ans + 1LL * val[j] * s % mod * n)%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * n)%mod;
                    sor=(sor + 1LL* val[j] *n)%mod;

                } else {
                    ans=(ans + 1LL * val[j] * s % mod  * cnt[j])%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * cnt[j])%mod;
                    sor=(sor + 1LL* val[j] *cnt[j])%mod;
                }
            }
            sor=sor*sor%mod;
            sum2=sum2 *(a[i]%mod) %mod;
            ans=(ans-sor)%mod;
            ans=(ans+sum2)%mod;
        }
        cout<<(ans+mod)%mod<<"\n";
    }

}

### 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
// Sometimes, the questions are complicated - and the answers are simple. //
#include<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld  long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int N=5e5+9, mod=1e9+7;;
int t, n, cnt[61], val[61];
ll a[N];
int main() {
    IO

    for(ll j=0 ; j<60 ; j++)
        val[j]=(1LL<<j)%mod;

    cin>>t;
    while(t--) {
        cin>>n;

        ll ans=0, s=0;
        memset(cnt,0,sizeof cnt);

        for(int i=1; i<=n; i++) {
            cin>>a[i];
            s=(s+a[i]%mod);
            for(ll j=0 ; j<60 ; j++)
                if(((1LL<<j)&a[i])!=0)
                    cnt[j]++;
        }

        s%=mod;

        for(int i=1; i<=n; i++) {
            ll sor=0;
            ll sum2=0;
            for(int j=0; j<60; j++) {
                if(((1LL<<j)&a[i])!=0) {

                    ans=(ans + 1LL * val[j] * s % mod * n)%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * n)%mod;
                    sor=(sor + 1LL* val[j] *n)%mod;

                } else {
                    ans=(ans + 1LL * val[j] * s % mod  * cnt[j])%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * cnt[j])%mod;
                    sor=(sor + 1LL* val[j] *cnt[j])%mod;
                }
            }
            sor=sor*sor%mod;
            sum2=sum2 *(a[i]%mod) %mod;
            ans=(ans-sor)%mod;
            ans=(ans+sum2)%mod;
        }
        cout<<(ans+mod)%mod<<"\n";
    }

}

```