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 rep(i,j,n) for(i=j;i<n;i++)
#define repi(i,j,n) for(i=j;i>n;i--)
#define inf 100001
#define M 1000000007
#define ll long long
#define vll vector<long long>
#define vi vector<int>
#define pll pair<long long,long long>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define vpll vector<pair<long long,long long> >
#define pb push_back
#define mp make_pair
#define F first
#define S second


ll power(ll a,ll m,ll mod){
	ll ans=1;
	while(m){
		if(m%2){
			ans*=a;
			ans%=(mod);
		}
		a=(a*a)%(mod);
		m>>=1;
	}
	return ans;
}


void init(long long* a,long long n,long long val){
	for(long long i=0;i<n;i++)
		a[i]=val;
}



int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);	
	ll t=1,i,j;
	ll n;
	cin>>n;
	ll a;
	ll c[62]={0};
	rep(i,0,n){
	    cin>>a;
	    for(j=1;j<=a;j=j<<1){
	        if(j&a)
	            c[(ll)log2(j)]++;
	    }
	}ll ans=0,k;
	rep(i,0,62){
	    k=c[i]*(n-c[i]);
	    ans+=k*pow(2,i);
	    ans%=M;
	}cout<<ans;
	return 0;
}