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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int MOD = 1000000007;
long long fact[1000005], finv[1000005], inv[1000005];
 
void f1(void) {
    fact[0] = fact[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < 1000005; i++){
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
long long f2(int n, int k) {
    int t;
    if (n < k) {
        t=k;
        k=n;
        n=t;
    };
    if (k==0) return 0;
    if (n < 0 || k < 0) return 0;
    return fact[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
int main(void){
    long long N, A[300005], a[65], S, p, i, j, k;
    scanf("%lld", &N);
    for (i=1; i<=N; i++) scanf("%lld", &A[i]);
    for (i=0; i<60; i++) a[i]=0;
    for (i=1; i<=N; i++) {
        for (j=0; j<60; j++) {
            p=1;
            for (k=1; k<=j; k++) p*=2;
            a[j]+=A[i]/p%2;
            // printf("%lld %lld\n", j, A[i]/p%2);
        }
    }
    // for (i=0; i<60; i++) printf("%lld ", a[i]);
    // printf("\n");
    S=0;
    f1();
    for (i=0; i<60; i++) {
        p=1;
        for (j=1; j<=i; j++) p*=2;
        p%=MOD;
        S+=p*a[i]*(N-a[i])%MOD;
        // printf("(%lld %lld)", a[i], N-a[i]);
    }
    printf("%lld", S);
    return 0;
}