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;
/**
********************************************************************
******************* Author:Bisnu sarkar ****************************
********************************************************************
**/
#define ull            unsigned long long
#define ll             long long
#define vi             vector<int>
#define pb             push_back
#define mp             make_pair
#define pii            pair<int,int>
#define vit            vector<int> :: iterator
#define sit            set<int> :: iterator
#define vrit           vector<int> :: reverse iterator
#define ff             first
#define ss             second
#define endl           '\n';
#define what_is(x)     cerr << #x << " is " << x << endl;
///*....Debugger....*/
#define error(args...) {vector <string> _v = split(#args, ','); err(_v.begin(), args); cout << endl;}
vector <string> split(const string &s, char c) {
    vector <string> v;
    stringstream ss(s); string x;
    while (getline(ss, x, c)) v.emplace_back(x);
    return move(v);
}
void err(vector <string> :: iterator it) {}
template <typename T, typename... Args>
void err(vector <string> :: iterator it, T a, Args... args) {
    cout << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << ", ";
    err(++it, args...);
}
 
int ini(){int x; scanf("%d",&x);return x;}
long long inl(){ long long x;scanf("%lld",&x); return x;}
double ind(){double x; scanf("%lf",&x);return x;}
#define Iin ini();
#define Lin inl();
#define Din ind();

int set_1(int n,int pos){return n = (n | (1<<pos));}
int reset_0(int n,int pos){return n= n & ~(1<<pos);}
bool check_bit(int n,int pos){return n = n & (1<<pos);}
bool check_bitl(ll n,int pos){return n = n & (1ll<<pos);}

int dx[8]= {1,0,-1,0,-1,-1,1,1};
int dy[8]= {0,1,0,-1,-1,1,-1,1};
const int N = (int) 1e6 + 5;
const ll M = (ll) 1e9 + 7;
const double pi=2* acos(0.0);
const double eps=1e-9;

ll v[64];

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL);  
    int t=1,te=0;
   
    while(t--){
        ll n;
        cin>>n;

        ll res=0ll;

        for(ll j=0;j<n;j++){
            ll x;
            cin>>x;
           // printf("sss\n");
            for(ll i=0;i<60;i++){
                int bit=(x>>i) & 1;
                if(bit){
                    res = (res+( (j-v[i]) * (1ll<<i)%M)%M)%M; 
                    v[i]++;
                }
                else{
                    res =  (res+(v[i]*(1ll<<i)%M)%M)%M;
                }


              //  printf("%lld %lld %lld\n",bit,v[i],res);
            }
        }

        printf("%lld\n", res%M);

        //printf("Case %d: ",++te);
    }
    return 0;
}