Problem:
Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author ExplodingFreeze
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define int long long
#define f(i,a,b) for(int i=(a);i<(b);i++)
const int MOD=1e9+7;
using pii=std::pair<int,int>;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

using namespace std;

int n,a[300005];
class DXorSum4 {
public:
	void solve(std::istream& in, std::ostream& out) {
		in >> n;
		f(i,0,n)
			in >> a[i];
		sort(a,a+n);
		int pow2[65]={0},cnt[65]={0};
		pow2[0]=1;
		f(i,1,62)
			pow2[i]=(pow2[i-1]*2)%MOD;
		f(j,0,62)
			if(a[0]&(1ll<<j))
				cnt[j]++;
		int ans=0;
		f(i,1,n)
		{
			f(j,0,62)
			{
				if(a[i]&(1ll<<j))
					ans+=((i-cnt[j])*pow2[j]);
				else
					ans+=(cnt[j]*pow2[j]);
				if(ans>=MOD)
					ans%=MOD;
			}
			f(j,0,62)
				if(a[i]&(1ll<<j))
					cnt[j]++;
		}
		out << ans << "\n";
	}
};


int32_t main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	DXorSum4 solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}