/*      Author : Kaustubh Gangwar
        Institute : IIITA        */
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define gcd __gcd
#define vi vector<int>
#define vll vector<long long>
#define all(x) (x).begin(), (x).end()
#define f(i, a, b) for (int i = a; i < b; i++)
#define fe(i, a, b) for (int i = a; i <= b; i++)
#define sz(x) (int)((x).size())
#define CODING ios_base::sync_with_stdio(false);
#define SAMRAT cin.tie(0), cout.tie(0);
#define mll map<ll, ll>
#define nl '\n'
#define lb lower_bound
#define ub upper_bound
typedef long long ll;
typedef pair<ll, ll> pll;
const int N = 200005;
ll bpow(ll a, ll b)
{
    a = a;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}
int32_t main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    CODING SAMRAT
    int test = 1;
    cin >> test;
    while (test--)
    {  
        ll n;
        cin>>n;
        ll a[n+1];
        ll bits[65]={0};
        fe(i,1,n)
        cin>>a[i];
        fe(i,1,n){
        fe(j,0,60){
        if((1LL<<j)&a[i])bits[j]++;
        }
        }

        ll b[n+5];
        fe(i,1,n)
        b[i]=a[i]+1;

        ll ans=0;

        fe(i,1,n){
            ll sam=0,rat=0;
        fe(j,0,60){
                ll temp=(1LL<<j)%mod;
                if((1ll<<j)&a[i]){
                    sam=(sam%mod+temp%mod*n%mod)%mod;
                    rat=(rat%mod+temp%mod*bits[j]%mod)%mod;
                }
                else {
                    sam=(sam%mod+temp%mod*bits[j]%mod)%mod;
                }
            }
           // cout<<ans<<nl;
            ans=(ans%mod+sam%mod*rat%mod)%mod;
        }

        cout<<ans%mod<<nl;
    }
    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
/*      Author : Kaustubh Gangwar
        Institute : IIITA        */
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define gcd __gcd
#define vi vector<int>
#define vll vector<long long>
#define all(x) (x).begin(), (x).end()
#define f(i, a, b) for (int i = a; i < b; i++)
#define fe(i, a, b) for (int i = a; i <= b; i++)
#define sz(x) (int)((x).size())
#define CODING ios_base::sync_with_stdio(false);
#define SAMRAT cin.tie(0), cout.tie(0);
#define mll map<ll, ll>
#define nl '\n'
#define lb lower_bound
#define ub upper_bound
typedef long long ll;
typedef pair<ll, ll> pll;
const int N = 200005;
ll bpow(ll a, ll b)
{
    a = a;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}
int32_t main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    CODING SAMRAT
    int test = 1;
    cin >> test;
    while (test--)
    {  
        ll n;
        cin>>n;
        ll a[n+1];
        ll bits[65]={0};
        fe(i,1,n)
        cin>>a[i];
        fe(i,1,n){
        fe(j,0,60){
        if((1LL<<j)&a[i])bits[j]++;
        }
        }

        ll b[n+5];
        fe(i,1,n)
        b[i]=a[i]+1;

        ll ans=0;

        fe(i,1,n){
            ll sam=0,rat=0;
        fe(j,0,60){
                ll temp=(1LL<<j)%mod;
                if((1ll<<j)&a[i]){
                    sam=(sam%mod+temp%mod*n%mod)%mod;
                    rat=(rat%mod+temp%mod*bits[j]%mod)%mod;
                }
                else {
                    sam=(sam%mod+temp%mod*bits[j]%mod)%mod;
                }
            }
           // cout<<ans<<nl;
            ans=(ans%mod+sam%mod*rat%mod)%mod;
        }

        cout<<ans%mod<<nl;
    }
    return 0;
}
```