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: Runtime Error)

#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pb push_back
#define f first
#define s second
#define MOD 1000000007
 
using namespace std;

uint64_t A[100001];

ll modexp(int i, ll p){
	if(p == 0) return 1;
	ll ans = modexp(i,p/2);
	ans = (ans*ans)%MOD;
	if(p%2) ans =  (i*ans)%MOD;
	return ans;
}

void solve(){
	int n; cin>>n;
	for(int i=0; i<n; i++){
		cin>>A[i];
	}
	ll ans=0;
	for(int bit=0; bit<60; bit++){
		ll x=0;
		for(int i=0; i<n; i++){
			if((A[i] & (1ULL<<bit))){
				x++;
			}
		}
		//cout<<bit<<" : "<<x<<" sex\n";
		x = (x*(n-x))%MOD;
		ans += (modexp(2,bit)*x)%MOD;
		ans = ans%MOD;
	}
	cout<<ans;

}


int main()
{ 
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout<<fixed<<setprecision(12);
	int t =1; //cin>>t;
	for(int i=1; i<=t; i++){
		solve();
		cout<<"\n";
	}
	return 0;

}