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 <string.h>
#include <stdlib.h>
#include <stdint.h> // uint64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

// <arr[0]> <arr[1]> .. <arr[size-1]>
int fget_array(int64_t *arr, int size) {
    int i;
    for(i = 0; i < size; i++) {
        scanf("%lld", &arr[i]);
    }
    return 0;
}

#define NUM_MAX 300000
#define POWER_MAX 60

#define power2(bit) (1LL<<bit)
#define DIVISOR 1000000007

int main(void) {
    int num = get_int();
    int64_t arr[NUM_MAX];
    static int cum[NUM_MAX+1][POWER_MAX+1];
    fget_array(arr, num);

    int i, bit;
    for(i = num-1; i >= 0; i--) {
        for(bit = 0; bit <= POWER_MAX; bit++) {
            if(!(arr[i] & power2(bit))) continue;
            cum[bit][i]++;
        }
        for(bit = 0; bit <= POWER_MAX; bit++) {
            cum[bit][i] += cum[bit][i+1];
        }
    }

    static int bit_sum[POWER_MAX+1];
    for(i = 0; i < num-1; i++) {
        for(bit = 0; bit <= POWER_MAX; bit++) {
            if(!(arr[i] & power2(bit))) {
                bit_sum[bit] += cum[bit][i+1];
            } else {
                bit_sum[bit] += ((num-i-1) - cum[bit][i+1]);
            }
        }
    }
    int64_t ans = 0;
    int64_t power2 = 1;
    for(i = 0; i <= POWER_MAX; i++) {
#ifdef DEBUG
        printf("%d: %d\n", i, bit_sum[i]);
#endif
        ans = (ans + power2 * bit_sum[i])%DIVISOR;
        power2 = (power2 * 2) % DIVISOR;
    }
    printf("%d\n", (int)ans);
    return 0;
}