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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>

#define dev 1000000007
long A_[300300];
long bubbleSort(long* A, long N)
{
	long flag=1;
	long i,tmp;
	while(flag){
		flag = 0;
		
		for(i=N-1;i>0;i--){
			if(A[i]<A[i-1]){
				tmp = A[i];
				A[i]=A[i-1];
				A[i-1]=tmp;
				flag=1;
//				cnt++;
			}
		}
	}
	return 0;
}
int main(){

	long N,max=-100;
	long i,j,dat;
	char s[110];
	
	scanf("%ld",&N);
	
	for(i=0;i<N;i++){
		scanf("%ld",&A_[i]);
		//A[dat]=1;
		//if(max<dat)max=dat;
	}
	//bubbleSort(A_, N);
	
	int cnt=0;
	long ans=0,tmp=-1,tmp2=0;
	for(i=0;i<N-1;i++){
			if(tmp!=A_[i]){
				tmp2=0;
			   // printf("in:%ld\n",A_[i]);
				for(j=i+1;j<N;j++){
//					if(A[j]==1){
						tmp2 +=A_[i]^A_[j];
						
						if(tmp2>dev)tmp2 %= dev;//printf("%ld:%ld %ld %ld\n",i,ans,A[i],A[j]);
//					}
				}
				tmp = A_[i];
				ans +=tmp2;
				if(ans>dev)ans %= dev;
			}
			else{
				ans +=tmp2;
				tmp = A_[i];
				if(ans>dev)ans %= dev;
			}
		//}

		
	}
	
	
	 printf("%ld\n",ans);
 return 0;
 }