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;
using ll=long long;
using vb=vector<bool>;
using vvb=vector<vb>;
using vd=vector<double>;
using vvd=vector<vd>;
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
using tll=tuple<ll,ll>;
using tlll=tuple<ll,ll,ll>;
using vs=vector<string>;

#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rep(i,n) range(i,0,n)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define range(i,a,n) for(int i=(a);i<(n);i++)

#define LINF ((ll)1ll<<60)
#define INF ((int)1<<30)
#define EPS (1e-9)
#define MOD (1000000007ll)
#define fcout(a) cout<<setprecision(a)<<fixed
#define fs first
#define sc second
#define PI 3.1415926535897932384

int dx[]={1,0,-1,0,1,-1,-1,1},dy[]={0,1,0,-1,1,1,-1,-1};

template<class T>bool chmax(T&a,T b){if(a<b){a=b; return true;}return false;}
template<class T>bool chmin(T&a,T b){if(a>b){a=b; return true;}return false;}
template<class S>S acm(vector<S>&a){return accumulate(all(a),S());}
template<class S>S max(vector<S>&a){return *max_element(all(a));}
template<class S>S min(vector<S>&a){return *min_element(all(a));}
void YN(bool b){cout<<(b?"YES":"NO")<<"\n";}
void Yn(bool b){cout<<(b?"Yes":"No")<<"\n";}
void yn(bool b){cout<<(b?"yes":"no")<<"\n";}
int sgn(const double&r){return (r>EPS)-(r<-EPS);} // a>0  : sgn(a)>0
int sgn(const double&a,const double&b){return sgn(a-b);} // b<=c : sgn(b,c)<=0

ll max(int a,ll b){return max((ll)a,b);} ll max(ll a,int b){return max(a,(ll)b);}
template<class T>void puta(T&&t){cout<<t<<"\n";}
template<class H,class...T>void puta(H&&h,T&&...t){cout<<h<<' ';puta(t...);}
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool a=1; for(auto s:t){os<<(a?"":" ")<<s;a=0;} return os;}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,ans=0;
	cin>>n;
	vl v(n);
	rep(i,n)cin>>v[i];
	vl cnt(100,0);
	rep(i,n){
		int j=0;
		ll tmp=v[i];
		while(tmp!=0ll){
			if(tmp&1ll)cnt[j]++;
			j++;
			tmp=tmp>>(1);
		}
	}
	rep(i,100){
		ll tmp=(((cnt[i]*(n-cnt[i]))%MOD)*(1ll<<i))%MOD;
		ans+=tmp;
		ans%=MOD;
	}
	puta(ans);
}