/**
 *	author:   fractal
 *	timus:    288481RF
 *	created: 12/30/20 22:57:50
**/

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

#define F first
#define S second 
#define mp make_pair
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define speed ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define sz(x) (int)x.size()
#define len(x) (int)strlen(x)
#define all(x) x.begin(), x.end()
#define debug cerr << "OK\n";
#define ub upper_bound
#define lb lower_bound 
#define nl printf("\n");
#define clbuff fflush(stdin);

mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef set <int> si;
typedef set <ll> sll;
typedef multiset <int> msi;
typedef multiset <ll> msll;
typedef map <int, int> mi;
typedef map <ll, int> mll;
 
const int N = 5e5 + 2;
const int M = 1e5;
const ll mod = 1e9 + 7;
const int inf = 2e9 + 3;
const ll INF = 1e18;

const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};

void files() {
	#ifndef PC
		freopen(".in", "r", stdin);
		freopen(".out", "w", stdout);
	#endif
}

ll T, n, x, was[61], pw[61], is[61][N], ans, a[N];

int main() {
	speed;
	cin >> T;
	pw[0] = 1;
	for (ll j = 1; j <= 60; ++j)
		pw[j] = (pw[j - 1] * 2LL) % mod;
	while (T--) {
		cin >> n;
		for (ll j = 60; j >= 0; --j) 
			was[j] = 0;
		for (int i = 1; i <= n; ++i) {
			cin >> x;
			a[i] = x;
			for (ll j = 60; j >= 0; --j) {
				is[j][i] = 0;
				if (x & (1LL << j)) {
					was[j]++;
					is[j][i] = 1;
				}
			}
		}
		ans = 0;
		for (int i = 1; i <= n; ++i) {
			ll mx = 0;
			a[i] %= mod;
			for (ll j = 0; j <= 60; ++j) {
				if (!is[j][i]) {
					mx = (mx + (was[j] * pw[j] % mod)) % mod;
				}
			}
			for (ll j = 0; j <= 60; ++j) {
				if (is[j][i]) {
					ans = (ans + (((was[j] * n % mod) * pw[j] % mod) * a[i] % mod)) % mod;
					ans = (ans + ((was[j] * pw[j] % mod) * mx % mod)) % mod;
				}
			}
		}
		
		cout << ans << '\n';
	}
}

### Prompt

Please create a solution in CPP 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:   fractal
 *	timus:    288481RF
 *	created: 12/30/20 22:57:50
**/

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

#define F first
#define S second 
#define mp make_pair
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define speed ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define sz(x) (int)x.size()
#define len(x) (int)strlen(x)
#define all(x) x.begin(), x.end()
#define debug cerr << "OK\n";
#define ub upper_bound
#define lb lower_bound 
#define nl printf("\n");
#define clbuff fflush(stdin);

mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef set <int> si;
typedef set <ll> sll;
typedef multiset <int> msi;
typedef multiset <ll> msll;
typedef map <int, int> mi;
typedef map <ll, int> mll;
 
const int N = 5e5 + 2;
const int M = 1e5;
const ll mod = 1e9 + 7;
const int inf = 2e9 + 3;
const ll INF = 1e18;

const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};

void files() {
	#ifndef PC
		freopen(".in", "r", stdin);
		freopen(".out", "w", stdout);
	#endif
}

ll T, n, x, was[61], pw[61], is[61][N], ans, a[N];

int main() {
	speed;
	cin >> T;
	pw[0] = 1;
	for (ll j = 1; j <= 60; ++j)
		pw[j] = (pw[j - 1] * 2LL) % mod;
	while (T--) {
		cin >> n;
		for (ll j = 60; j >= 0; --j) 
			was[j] = 0;
		for (int i = 1; i <= n; ++i) {
			cin >> x;
			a[i] = x;
			for (ll j = 60; j >= 0; --j) {
				is[j][i] = 0;
				if (x & (1LL << j)) {
					was[j]++;
					is[j][i] = 1;
				}
			}
		}
		ans = 0;
		for (int i = 1; i <= n; ++i) {
			ll mx = 0;
			a[i] %= mod;
			for (ll j = 0; j <= 60; ++j) {
				if (!is[j][i]) {
					mx = (mx + (was[j] * pw[j] % mod)) % mod;
				}
			}
			for (ll j = 0; j <= 60; ++j) {
				if (is[j][i]) {
					ans = (ans + (((was[j] * n % mod) * pw[j] % mod) * a[i] % mod)) % mod;
					ans = (ans + ((was[j] * pw[j] % mod) * mx % mod)) % mod;
				}
			}
		}
		
		cout << ans << '\n';
	}
}

```