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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//include
//------------------------------------------
#include <bits/stdc++.h>
using namespace std;
//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> P;
typedef long long ll;
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
#define INF (ll)1000000007
// chmax chmin
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

long long add_modp(long long a, long long b, long long p){
    // a+b mod p
    return (a % p + b % p) % p;
}

long long mul_modp(long long a, long long b, long long p){
    // a*b mod p
    return ((a % p) * (b % p)) % p;
}

long long pow_modp(long long a, long long n, long long p){
    // a^n mod p
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = mul_modp(res, a, p);
        a = mul_modp(a, a, p);
        n >>= 1;
    }
    return res;
}

int main(void){
    int N;
    cin >> N;
    ll A[N];
    REP(i,N) cin >> A[i];

    VI bit(60,0);
    REP(i,N){
        ll A_bit = A[i];
        int i_bit = 0;
        while(A_bit){
            if(A_bit & 1) bit[i_bit]++;
            i_bit++;
            A_bit >>= 1;
        }
    }

    VI num(60);
    REP(i,60) num[i] = bit[i]*(N-bit[i]);

    ll ans = 0;
    ll pow = 1;
    REP(i,60){
        ans = add_modp(ans, mul_modp(num[i], pow, INF), INF);
        pow = mul_modp(pow, 2, INF);
    }

    cout << ans << endl;

    return 0;
}