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: Time Limit Exceeded)

#include <bits/stdc++.h>
struct Fast {Fast(){std::cin.tie(0);std::ios::sync_with_stdio(false);}} fast;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < (int)(n); i++)
#define repR(i, n) for(ll i = n; i >= 0; i--)
#define For(i, m, n) for(ll i = m; i < n; i++)

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) v.begin(), v.end()

using namespace std;

ll mod = 1e9+7;
inline ll gcd(ll a, ll b){
    ll r = a % b;
    while(r!=0){
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

inline ll lcm(ll x, ll y)
{
    if(x == 0 || y == 0) {
        return -1;
    }
    
    return (x * y / gcd(x, y));
}

inline ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

inline ll nCr(ll n, ll r) {
    if ( r * 2 > n ) r = n - r;
    ll dividend = 1;
    ll divisor  = 1;
    for ( ll i = 1; i <= r; ++i ) {
        dividend *= (n-i+1);
        divisor  *= i;
    }
    return (dividend / divisor);
}

// 因数分解
inline vector<pair<ll, int>>factorize(ll n){
    vector<pair<ll, int>> res;
    for(ll i = 2; i * i <= n; ++i){
        if(n % i) continue;
        res.emplace_back(i, 0);
        while(n % i == 0){
            n /= i;
            res.back().second++;
        }
    }
    if(n != 1) res.emplace_back(n, 1);
    return res;
}

inline bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}

//階乗
inline ll factorialMethod(ll k){
    ll sum = 1;
    for (int i = 1; i <= k; ++i)
    {
        sum *= i;
    }
    return sum;
}

inline ll vector_finder(std::vector<ll> vec, ll number) {
    auto itr = std::find(vec.begin(), vec.end(), number);
    size_t index = std::distance( vec.begin(), itr );
    if (index != vec.size()) { // 発見できたとき
        return 1;
    }
    else { // 発見できなかったとき
        return 0;
    }
}

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    
    rep(i, n){
        cin >> a[i];
    }
    
    
    ll ans = 0;
    rep(i, n-1){
        if(a[i] == 1){
            For(j, i+1, n){
                if(a[j] % 2 == 0){
                    ans += a[j]+1;
                }else{
                    ans += a[j]-1;
                }
            }
            continue;
        }
        
        For(j, i+1, n){
            ans += a[i] ^ a[j];
            ans %= mod;
        }
    }
    
    cout << ans << endl;
    return 0;
    
}