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//TLEの原因になるので注意！！！！！！！！！！！
#include<cmath>
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<int> arr;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
#define mins(x,y) x=min(x,y)
#define maxs(x,y) x=max(x,y)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;
static const int INF = 1000000007;
#define mp make_pair
#define sz(x) int(x.xise())
#define mod 1000000007
#define reps(i,s,n) for(int i = s; i <= n; i++)
deque<int> deq;
//#define int long long

//cerrはコメントアウトしなくても通る//ただしTLEを引き起こすことがある

ll ans=0;
ll a[300005];
ll b[65];
ll f[65];

int main(){
    f[0]=1;
    rep2(i,61){
        f[i]=f[i-1]*2;
    }
    ll n;
    cin >> n;
    rep2(i,n){
        ll a;
        cin >> a;
        ll k=0;
        while(a>0){
            if(a%2==1){
                b[k]++;
            }
            a/=2;
            k++;
        }
    }
    //cerr<<'i'<<endl;
    rep(i,61){
        ans+=f[i]*b[i]*(n-b[i]);
        ans%=mod;
    }
    cout<<ans<<endl;
    return 0;
}