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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int 		  long long
#define	pb 			  push_back
#define all(a)		  (a).begin(),(a).end()
#define sz(x)		  (int)x.size()
#define endl	 	  "\n"
#define MOD 		  1000000007
#define FOR(i, a, b)  for(int i=a; i<b; i++)
//int dp[1000001] = {0};
void solve(){
	int n, res=0;
	cin>>n;
	int arr[n];
	FOR(i, 0, n) cin>>arr[i];
	for(int b = 0; b <= 60; b++){
        int x[2] = {0};
        for(int i = 0; i < n;i++){
            x[(arr[i]>>b)&1]++;
        }
        res += (x[0]*x[1])%MOD*((1LL<<b)%MOD)%MOD;
        res %= MOD;
    }
    cout<<res;
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t=1;
	//cin>>t;
	FOR(i, 1, t+1){	
	//	cout<<"Case #"<<i<<":"<<endl;
		solve();
	}
	return 0;
 
}