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>
using namespace std;
#define ll  long long int
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define MOD 1000000007
#define FAST ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);srand(time(NULL));
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 
const int N = 100005;
int main()
{
  
   FAST;
   ll n;
   cin >> n;
   ll a[n];
   ll sum = 0;
   for(ll i=0;i<n;i++)
   {
       cin >> a[i];
       sum += a[i];
   }
   ll ans = 0;
   ans = (n-1)*sum;
   ans %= MOD;
   ll temp=1;
   for(ll i=0;i<=60;i++)
   {
       ll cnt = 0;
       for(ll j=0;j<n;j++)
       {
           if(temp & a[j])  cnt++;
       }
       cnt = (cnt*(cnt-1))/2;
       cnt %= MOD;
       ans = (ans - (((cnt)%MOD*(temp)%MOD)*2)%MOD +MOD)%MOD;
       temp *= 2;
   }
   cout<<ans;
}