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 <bits/stdc++.h>
using namespace std;

#define fst first
#define scd second
#define PB push_back
#define MP make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define omajinai ios::sync_with_stdio(false);cin.tie(0)
#define rep(i,x) for(int i=0;i<(int)(x);++i)
#define rep1(i,x) for(int i=1;i<=(int)(x);++i)
#define rrep(i,x) for(int i=(int)x-1;i>=0;--i)
#define rrep1(i,x) for(int i=(int)x;i>=1;--i)

using ll=long long;
using ld=long double;
using vi=vector<int>;
using vvi=vector<vi>;
using pii=pair<int, int>;
using vpii=vector<pii>;

template<class T,class U>ostream&operator<<(ostream&os,const pair<T,U>p){os<<"("<<p.fst<<", "<<p.scd<<")";return os;}
template<class T>ostream&operator<<(ostream&os,const vector<T>v){rep(i,v.size()){if(i)os<<", ";os<<v[i];}return os;}
template<typename T>T&max(T&a,T&b){if(a>=b)return a;return b;}
template<typename T>T&min(T&a,T&b){if(a < b)return a;return b;}
template<typename T>bool chmax(T&a,T b){if(a < b){a=b;return true;}return false;}
template<typename T>bool chmin(T&a,T b){if(a > b){a=b;return true;}return false;}

constexpr int inf = 1e9;
constexpr ll linf = 3e18;
constexpr double eps = 1e-9;

const int mo = 1e9+7;

using Int = long long;

int N;
int counter[65]; // counter[i] := 右からi bit目の1の数

signed main()
{
    cin >> N;

    rep(i, N) {
        Int A; cin >> A;
        rep(j, 63) {
            if ((A >> j) & 1) counter[j]++;
        }
    }

    Int ans = 0;

    rep(i, 63) {
        ans += (1ll << i) * counter[i] * (N - counter[i]);
        ans %= mo;
    }

    cout << ans << endl;
}