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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> l_l;
typedef pair<int,int> i_i;
const ll MOD=1e9+7;
const int INF=1<<30;

const int mod=1000000007;
struct mint{
  ll x;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const{return mint(-x);}
  mint& operator+=(const mint a){
    if((x+=a.x)>=mod) x-=mod;
    return *this;
  }
  mint& operator-=(const mint a){
    if((x+=mod-a.x)>=mod) x-=mod;
    return *this;
  }
  mint& operator*=(const mint a){
    (x*=a.x)%=mod;
    return *this;
  }
  mint operator+(const mint a) const{
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const{
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const{
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const{
    if(!t) return 1;
    mint a=pow(t>>1);
    a*=a;
    if(t&1) a*=*this;
    return a;
  }

  // for prime mod
  mint inv() const{
    return pow(mod-2);
  }
  mint& operator/=(const mint a){
    return (*this)*=a.inv();
  }
  mint operator/(const mint a) const{
    mint res(*this);
    return res/=a;
  }
};

ll c[60];

int main(){
  int N; cin >>N;
  vector<ll> a(N);
  rep(i,N) cin >> a[i];
  rep(i,N){
    for(int j=0;j<60;j++){
      if(a[i]&(1ll<<j)) c[j]++;
    }
  }
  mint ans=0;
  rep(j,60){
    ans+=(1ll<<j)%MOD*c[j]*(N-c[j]);
  }
  cout << ans.x << endl;
}