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: Accepted)

char buf[10000000],out[11];
sz;
cnt[60];
main()
{
	read(0,buf,10000000);
	long N=0;
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		N=N*10+k-'0';
	}
	for(int c=N;c--;)
	{
		long A=0;
		for(;;)
		{
			char k=buf[sz++];
			if(k<'0')break;
			A=A*10+k-'0';
		}
		for(int i=0;A;i++)
		{
			cnt[i]+=A&1;
			A>>=1;
		}
	}
	int ans=0;
	for(int i=59;i>=0;i--)
	{
		ans=(ans*2+(N-cnt[i])*cnt[i])%1000000007;
	}
	int outsz=10;
	do{
		out[--outsz]=ans%10+'0';
	}while(ans/=10);
	write(1,out+outsz,10-outsz);
}