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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const ll mod = 1e9 + 7;
ll modpow(ll A, ll B) {
	ll kotae = 1;
	while (B > 0) {
		if (B & 1) kotae = kotae * A % mod;
		A = A * A % mod;
		B >>= 1;
	}
	return kotae;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int N;
	cin >> N;
	ll A[300000];
	rep(i, N) cin >> A[i];

	int kazu[60] = {};
	rep(i, N) {
		rep(j, 60) {
			kazu[j] += A[i] >> j & 1;
		}
	}

	ll kotae[60] = {};
	rep(i, N) {
		rep(j, 60) {
			kazu[j] -= A[i] >> j & 1;
		}

		rep(j, 60) {
			if (A[i] >> j & 1) {
				kotae[j] += N - 1 - kazu[j];
			}
			else {
				kotae[j] += kazu[j];
			}
		}

		rep(j, 60) {
			kazu[j] += A[i] >> j & 1;
		}
	}

	ll ans = 0;
	rep(i, 60) {
		kotae[i] %= mod;
		ll tmp = (1ll << i) % mod;
		ans += kotae[i] * tmp % mod;
	}
	co(ans * modpow(2, mod - 2) % mod);


	Would you please return 0;
}