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"
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair <ll, ll > pll;

//bool comp(
//
//	st pll &a, const pll &b) {
//	return a.first < b.first;
//}
//std::vector<int> enum_div(int n)//nの約数を列挙
//{
//	std::vector<int> ret;
//	for (int i = 1; i*i <= n; ++i)
//	{
//		if (n%i == 0)
//		{
//			ret.push_back(i);
//			if (i != 1 && i*i != n)
//			{
//				ret.push_back(n / i);
//			}
//		}
//	}
//	return ret;
//}


template<int N>
int factorial() {
	return factorial<N - 1>() * N;
}

template<>  /// 特殊化：factorial<N=”1”>だったら，値を返す
int factorial<1>() {
	return 1;
}
int main()
{
	priority_queue<int> q;
	string s = {};
	ll h = 0, w = 0, k = 0, a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0, ans = 0;
	ll x[300005] = {};
	ll y[620] = {};
	ll y2[302][302] = {};
	cin >> a;
	for (ll i = 1; i <= a; i++)
	{
		cin >> x[i];
	}
	for (ll i = 1; i <= a; i++)
	{
		b = x[i];
		c = 0;
		while (b > 0)
		{
			if (b % 2 == 1)
			{
				y[c]++;
			}
			b /= 2;
			c++;
		}
	}
	//for (int i = 0; i < 60; i++)
	//{
	//	cout << y[i] << endl;

	//}
	//cout << endl;

	for (ll i = 0; i <= 62; i++)
	{
		ans += (y[i] * (a - y[i])*(ll(pow(2, i)) % 1000000007))% 1000000007;
		//cout << ans << endl;
		//cout << (ll(pow(2, i)) % 1000000007) << endl;
		//cout << endl;
		ans = ans % ll(1000000007);

	}

	cout << ans << endl;

	//for (int i = 1; i < a; i++)
	//{
	//	for (int j = i+1; j <= a; j++)
	//	{
	//		ans += x[i] ^ x[j];
	//		ans %= 1000000007;
	//	}
	//}



	//cout<<factorial<10><<endl;

	//cin >> a >> b;
	//for (int i = 0; i < a; i++)
	//{
	//	cin >> s[i];
	//	if (s[i] >= b)
	//	{
	//		ans++;
	//	}
	//}

	//cout << ans << endl;
	return 0;
}
//2つの組の数をソートしたりするやつ
//vector<pair<ll, ll> > hoges(a);
//for (int i = 1; i <= a; i++)
//{
//	cin >> h;
//	hoges[i - 1] = make_pair(h, i);
//}
//sort(hoges.begin(), hoges.end(), comp);
//for (int i = 0; i < a; i++)
//{
//	cout << hoges[i].second << endl;
//}