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)

#define _GLIBCXX_DEBUG
typedef long long ll;
#include <bits/stdc++.h>

using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

ll modpow(ll a, ll n, ll mod)
{
	ll res = 1;
	while (0 < n)
	{
		if (n & 1)
		{
			res = res * a % mod;
		}
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

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

	ll ans = 0;
	for (int bit = 0; bit < 60; bit++)
	{
		ll cnt0 = 0;
		ll cnt1 = 0;
		ll tmp = (1LL << bit);
		for (int i = 0; i < n; i++)
		{
			if ((1LL << bit) & a[i])
			{
				cnt1++;
			}
			else
			{
				cnt0++;
			}
		}
		tmp *= cnt0 * cnt1;
		ans += tmp;
	}
	cout << ans << endl;
}