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: Accepted)

/*author: hyperion_1724
  date: 
*/

//Optional FAST

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")

//Required Libraries

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>

//Required namespaces

using namespace std;
using namespace __gnu_pbds;

//Required defines

#define endl '\n'

#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define trav(x,A) for(auto &x:A)
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)

//Required typedefs

typedef tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<long long,null_type,greater<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set1;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
typedef vector<int> veci;
typedef vector<long long> vecl;

//Required Constants

const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);

//Required Functions

ll power(ll x,ll y) 
{ 
    if (y == 0) 
        return 1; 
    ll p = power(x, y/2) % MOD; 
    p = (p * p) % MOD; 
  
    return (y%2 == 0)? p : (x * p) % MOD; 
}
ll modInverse(ll a) 
{ 
   return power(a,MOD-2); 
   
}

//Work

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

    READV(N);
    ll A[61]={0};
    FORI(i,0,N)
    {
        READV(x);
        FORI(j,0,61)
        {
            if (x%2==1)
            {
                A[j]++;
            }
            x=x/2;
        }
    }
    ll ans=0;
    FORI(i,0,61)
    {
        ll z=((N-A[i])*A[i])%MOD;
        z=(z*power(2,i))%MOD;
        ans=(ans+z)%MOD;
    }
    cout<<ans<<endl;

    return 0;
}