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 <ext/pb_ds/assoc_container.hpp>
#define F first
#define S second
#define PB push_back
#define MAXN 100005
#define MP make_pair
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define debug(x) cout << '>' << #x << ':' << x << endl;
#define SQ(a) (a)*(a)
#define nl "\n"
#define INF std::numeric_limits<int>::max()
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_set;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
ll mod=1e9+7;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	ll n;
	cin>>n;
	ll a[n] = {0};
	ll cmn_cnt[63] = {0};
	for(ll i = 0 ; i < n ; i++) {
		cin>>a[i];
	}
	ll ans = 0;
	for(ll i = 0; i <= 60;i++){
		ll cnt=0;
		for(int j = 0; j < n; j++){
				cnt+=(a[j]>>i)&1;
		}
		int temp =1;
		rep(j,0,i-1)temp*=2;
		ans=(ans+(((temp*cnt)%mod)*(n-cnt))%mod)%mod;
	}
	cout<<ans%mod<<nl;
	return 0;
}