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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;

public class Program
{
    private int N;
    private long[] A;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        A = sc.LongArray();

        // [i+1,N)のxor

        // [0,i)で j桁目が1の奴の個数の和 cnt[j,i]
        long[][] cnt = new long[60][];
        for (int i = 0; i < 60; i++)
        {
            cnt[i] = new long[N + 1];
        }

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < 60; j++)
            {
                long bit = 1L << j;
                if ((A[i] & bit) > 0)
                {
                    cnt[j][i + 1] = cnt[j][i] + 1;
                }
                else
                {
                    cnt[j][i + 1] = cnt[j][i];
                }
            }
        }

        ModInt ans = 0;

        for (int i = 0; i < N - 1; i++)
        {
            for (int j = 0; j < 60; j++)
            {
                // [i+1, N)の個数
                long on = cnt[j][N] - cnt[j][i + 1];
                long off = N - i - 1 - on;

                long b = 1L << j;
                bool bit = (A[i] & b) > 0;
                if (bit)
                {
                    ans += off * b;
                }
                else
                {
                    ans += on * b;
                }
            }
        }
        
        Console.WriteLine(ans);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

#region ModInt

/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt
{
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = (int) 1e9 + 7;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;

    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n)
    {
        num = n;
    }

    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString()
    {
        return num.ToString();
    }

    public static ModInt operator +(ModInt l, ModInt r)
    {
        l.num += r.num;
        if (l.num >= Mod) l.num -= Mod;
        return l;
    }

    public static ModInt operator -(ModInt l, ModInt r)
    {
        l.num -= r.num;
        if (l.num < 0) l.num += Mod;
        return l;
    }

    public static ModInt operator *(ModInt l, ModInt r)
    {
        return new ModInt(l.num * r.num % Mod);
    }

    public static implicit operator ModInt(long n)
    {
        n %= Mod;
        if (n < 0) n += Mod;
        return new ModInt(n);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k)
    {
        return Pow(v.num, k);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k)
    {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1)
                ret = ret * v % Mod;
        return new ModInt(ret);
    }

    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v)
    {
        return Pow(v, Mod - 2);
    }
}

#endregion

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}