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: Wrong Answer)

#include<stdio.h>
#include<math.h>

#define MOD 1000000007

long long int modepow(long long int a, long long int n){
    long long int res = 1;
    while(n>0){
        if(n&1){
            res = res*a % MOD;
        }
        a = a*a % MOD;
        n >>= 1;
    }
    return res;
}

int main(){
    int n;
    scanf("%d", &n);
    long long int a[n];
    int cnt[63];
    for(int i=0; i<63; i++){
        cnt[i] = 0;
    }
    for(int i=0; i<n; i++){
        long long int buf;
        scanf("%lld", &buf);
        a[i] = buf;
        int c=0;
        while(buf > 0){
            if(buf % 2){
                cnt[c]++;
            }
            buf = buf/2;
            c++;
        }
    }
    long long int ans = 0;
    for(long long int i=0; i<=61; i++){
        int x = cnt[i], y = n-cnt[i];
        ans += (((x*y)%MOD)*(modepow(2, i) % MOD))%MOD;
        ans = ans % MOD;
    }
    printf("%lld", ans);

    return 0;
}