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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL << 60)
#define MOD 1000000007
#define PI 3.14159265358979323846

#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORR(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define SZ(x) ((ll)(x).size())
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) ASC((x)); REV((x))
#define pb push_back
#define eb emplace_back


ll modpow(ll a, ll n, ll m)
{
    if(n == 0) return 1;
    ll t = modpow(a, n >> 1, m);
    t = t * t % m;
    if(n & 1) t = t * a % m;
    return t; 
}

ll modinv(ll a, ll m)
{
    return modpow(a, m - 2, m);
}

int main()
{
    ll N;
    cin >> N;
    vec A(N);
    REP(i,0,N) cin >> A[i];

    vec C(60, 0);
    REP(i,0,N) {
        REP(j,0,60) C[j] += ((A[i] >> j) & 1);
    }

    ll sum = 0;
    REP(i,0,N) {
        REP(j,0,60) {
            if((A[i] >> j) & 1) sum = (sum + ((1LL << j) % MOD) * (N - C[j])) % MOD;
            else sum = (sum + ((1LL << j) % MOD) * C[j]) % MOD;
        }
    }
    PR((sum*modinv(2,MOD))%MOD);

    return 0;
}

/*



*/