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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function Main(input) {
    input = input.trim().split("\n").map(function(x) { return x.split(" ")});    
    var C = 1e+9 +7;
    var C2 = 1e+5;
    var N = parseInt(input[0][0], 10);
    var vec_A = input[1];
    var bitAlist = [];
    var v_ruijo = [1n];
    var v_ruijoR = [1];
    for (var i = 0; i < 59; i++){
        v_ruijo.push(v_ruijo[i] * 2n);
        v_ruijoR.push(v_ruijoR[i] * 2 % C);
    }
    var tmp, tmp2;
    for (var i = 0; i < N; i++){
        bitAlist.push([]);
        if (BigInt(vec_A[i]) < v_ruijo[53]){
            tmp = parseInt(vec_A[i]).toString(2);
            tmp = ('0'.repeat(60) + tmp).substr(-60, 60);
            bitAlist[i] = tmp.split("").map(e => parseInt(e, 10));
        } else {
            tmp2 = BigInt(vec_A[i]);
            tmp = '';
            for (var j = 0; j < (60 - 53); j++){
                if (tmp2 >= v_ruijo[59 - j]){
                    tmp2 = tmp2 - v_ruijo[59 - j];
                    tmp += '1';
                } else {
                    tmp += '0';
                }
            }
            tmp += ('0'.repeat(53) + tmp2.toString(2)).substr(-53, 53);
            bitAlist[i] = tmp.split("").map(e => parseInt(e, 10));
        }
    }
    var ans = 0;
    var nZero, nOne;
    for (var i = 0; i < 60; i++){
        nZero = nOne = 0;
        for(var j = 0; j < N; j++){
            if (bitAlist[j][i] === 1) nOne++;
            else nZero++;
        }
        ans = ans + kakeruR(v_ruijoR[59 - i], (nOne * nZero % C), C, C2);
        ans %= C;
    }
    console.log(ans);
}


function kakeruR(a, b, C, C2){
    var tmpQ, tmpR, ans;
    tmpQ = (b / C2 | 0);
    tmpR = b - (tmpQ * C2);
    ans = (a * tmpQ % C) * C2 % C + (a * tmpR);
    ans %= C;
    return ans;
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));