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)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
#define rep(i,a,b) for(ll i=a ; i<b ; i++)
#define qrep(que, ite) for(auto ite=begin(que) ; ite!=end(que) ; ite++)
const int max_n = 1e5;
const ll mod = 1e9+7;
const ll INF = 1LL<<60;
const int inf = 1e5;
//typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
ll mo = 998244353;
struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) { }
    void init(int n) { par.assign(n, -1); }

    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x) {
        return -par[root(x)];
    }
};
int main(){
  ll n;  cin >> n;
  vector<ll> a(n);
  ll cnt[60]={0};
  rep(i,0,n){
    ll add; cin >> add;
    rep(j,0,60){
      if(add & (1LL<<j)){
        cnt[j]++;
      }
    }
  }
  ll ans = 0;
  rep(i,0,60){
    ans += (1LL<<i)%mod * cnt[i] % mod * (n-cnt[i]) % mod;
    ans %= mod;
  }
  cout << ans << endl;

  return 0;
}