#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;

#define N 600100
#define MOD 1000000007  //998244353
#define ll long long
#define rep(i, n) for(ll i = 0; i < n; ++i)
#define rep2(i, a, b) for(ll i = a; i <= b; ++i)
#define rep3(i, a, b) for(ll i = a; i >= b; --i)
#define eb emplace_back
#define pb push_back
#define all(c) (c).begin(), (c).end()
#define vi vector<int>
#define pii pair<int,int>
#define pll pair<ll,ll>


int main() {
	ll c[62];
	ll p2[62];
	ll a[N];
	ll s, ans, x, y, z, n, ss;
	p2[0] = (ll)1;
	rep(i, 61)p2[i + 1] = (p2[i] * 2) % MOD;
	ll t;
	cin >> t;
	rep(tt, t) {
		cin >> n;
		rep(i, 60)c[i] = 0;
		ss = 0;
		rep(i, n) {
			cin >> a[i];
			ss = (ss + a[i]) % MOD;
			x = a[i];
			rep(j, 60) {
				if (x % 2 == 1)c[j]++;
				x /= 2;
			}
		}
		ans = 0;
		rep(i, n) {
			x = a[i];
			s = 0;
			rep(j, 60) {
				if (x % 2 == 1) {
					y = (c[j] * p2[j]) % MOD;
					s = (s + y) % MOD;
				}
				x /= 2;
			}
			x = a[i] % MOD;
			x = (x*n) % MOD;
			x = (x*s) % MOD;
			ans = (ans + x) % MOD;
			x = (ss*s) % MOD;
			ans = (ans + x) % MOD;
			x = (s*s) % MOD;
			ans = (ans + MOD - x) % MOD;
		}
		cout << ans << endl;
	}

	return 0;
}

### Prompt

Create a solution in cpp for 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
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;

#define N 600100
#define MOD 1000000007  //998244353
#define ll long long
#define rep(i, n) for(ll i = 0; i < n; ++i)
#define rep2(i, a, b) for(ll i = a; i <= b; ++i)
#define rep3(i, a, b) for(ll i = a; i >= b; --i)
#define eb emplace_back
#define pb push_back
#define all(c) (c).begin(), (c).end()
#define vi vector<int>
#define pii pair<int,int>
#define pll pair<ll,ll>


int main() {
	ll c[62];
	ll p2[62];
	ll a[N];
	ll s, ans, x, y, z, n, ss;
	p2[0] = (ll)1;
	rep(i, 61)p2[i + 1] = (p2[i] * 2) % MOD;
	ll t;
	cin >> t;
	rep(tt, t) {
		cin >> n;
		rep(i, 60)c[i] = 0;
		ss = 0;
		rep(i, n) {
			cin >> a[i];
			ss = (ss + a[i]) % MOD;
			x = a[i];
			rep(j, 60) {
				if (x % 2 == 1)c[j]++;
				x /= 2;
			}
		}
		ans = 0;
		rep(i, n) {
			x = a[i];
			s = 0;
			rep(j, 60) {
				if (x % 2 == 1) {
					y = (c[j] * p2[j]) % MOD;
					s = (s + y) % MOD;
				}
				x /= 2;
			}
			x = a[i] % MOD;
			x = (x*n) % MOD;
			x = (x*s) % MOD;
			ans = (ans + x) % MOD;
			x = (ss*s) % MOD;
			ans = (ans + x) % MOD;
			x = (s*s) % MOD;
			ans = (ans + MOD - x) % MOD;
		}
		cout << ans << endl;
	}

	return 0;
}

```