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;

typedef long long int ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<string>vst;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef map<int,int> dct;

#define fastio ios_base::sync_with_stdio(false);cin.tie(0)
#define all(ct) ct.begin() , ct.end()
#define endl "\n"
#define fr(i,a) for(auto i:a)
#define f(i,a,b) for(int i=a;i<b;++i)
#define fd(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define in(d,v) d.find(v)!=d.end()
#define mp make_pair
#define pi 3.14159265

bool compare(pll a, pll b){
    return (a.first*b.second) < (a.second*b.first);
}

const int mod = 1e9 + 7;
const ll inf = ll(1e18);

ll modexp(ll x,ll n){
    if(n==0)return 1;
    if(n%2==0)return modexp((x*x)%mod,n/2);
    return (x*modexp((x*x)%mod,n/2))%mod;
}

int main(){
    fastio;
    int n;
    cin>>n;
    vl a(n,0);
    f(i,0,n)cin>>a[i];
    ll ans = 0;
    f(i,0,60){
        int sb = 0, usb = 0;
        ll temp = 0;
        f(j,0,n){
            if (a[j] % 2 == 0)usb++;
            else sb++;
            a[j] /= 2;
        }
        temp = (sb * usb)%mod;
        temp=(temp* ll(pow(2,i)))%mod;
        ans = (ans+temp)%mod;
    }
    cout<<ans<<endl;
}