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)

/*
これを入れて実行
g++ code.cpp
./a.out
*/

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>
#include <stack>

using namespace std;
typedef long long ll;
typedef long double ld;

int dy4[4] = { -1, 0, +1, 0 };
int dx4[4] = { 0, +1, 0, -1 };
int dy8[8] = { -1, -1, 0, 1, 1, 1, 0, -1 };
int dx8[8] = { 0, 1, 1, 1, 0, -1, -1, -1 };

const long long INF = 1LL << 60;
const ll MOD = 1e9 + 7;

bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
	return f.second > s.second;
}

ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}

ll lcm(ll a, ll b){
	return a / gcd(a, b) * b;
}

ll nCr(ll n, ll r){
	if (r == 0 || r == n){
		return 1;
	}
	else if (r == 1){
		return n;
	}
	return (nCr(n - 1, r) + nCr(n - 1, r - 1));
}

ll nPr(ll n, ll r){
	r = n - r;
	ll ret = 1;
	for (ll i = n; i >= r + 1; i--) ret *= i;
	return ret;
}

//-----------------------ここから-----------

int main(void){
	int n;
	cin >> n;

	vector<ll> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];

	ll ans = 0;

	for (ll i = 0; i < 60; i++){
		ll cntzero = 0;
		ll cntone = 0;

		for (int j = 0; j < n; j++){
			if ((a[j] >> i) & 1 == 1){
				cntone++;
			}
			else {
				cntzero++;
			}
		}

		ll tmp = ((cntone * cntzero) % MOD) * ((ll)pow(2, i) % MOD);
		tmp %= MOD;

		ans += tmp;
		ans %= MOD;

	}

	cout << ans << endl;
}