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)

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store")  
#pragma GCC optimize ("-fno-defer-pop")
typedef long long int ll; 
typedef long double ld; 

ll mod = 1000000007;

ll power(ll a,ll b)
{
	ll r = 1;
	while(b>0)
	{
		if(b%2==1)
		{
			r = (mod+(r*a)%mod)%mod;
		}
		a = ((a*a)%mod+mod)%mod;
		b = b/2;
	}
	return r;
}

int main()
{
	std::ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	ll n;
	cin>>n;
	
	ll arr[n];
	
	for(ll i=0;i<n;i++)
	{
		cin>>arr[i];
	}
	
	ll count = 0;
	ll i = 0;
	ll ans = 0;
	while(1)
	{
		ll k=0;
		ll b = 0;
		if(i<60){
			
			for(ll j=0;j<n;j++)
			{
				if(((arr[j]>>i)&1)==1)
				{
					k++;
				}
				else{
					b++;
				}
			}
		}
		count+=k*b;
		if(count%2==1)
		{
			ans = (mod+(ans+power(2,i))%mod)%mod;
		}
		count = count/2;
		if(count==0 and i>=60)break;
		i++;
	}
	cout<<ans<<endl;
	
	return 0;
}