Problem:
Score : 400 points

Problem Statement

We have N balls. The i-th ball has an integer A_i written on it.

For each k=1, 2, ..., N, solve the following problem and print the answer.

Find the number of ways to choose two distinct balls (disregarding order) from the N-1 balls other than the k-th ball so that the integers written on them are equal.

Constraints

3 \leq N \leq 2 \times 10^5

1 \leq A_i \leq N

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

For each k=1,2,...,N, print a line containing the answer.

Sample Input 1

5
1 1 2 1 2

Sample Output 1

2
2
3
2
3

Consider the case k=1 for example. The numbers written on the remaining balls are 1,2,1,2.

From these balls, there are two ways to choose two distinct balls so that the integers written on them are equal.

Thus, the answer for k=1 is 2.

Sample Input 2

4
1 2 3 4

Sample Output 2

0
0
0
0

No two balls have equal numbers written on them.

Sample Input 3

5
3 3 3 3 3

Sample Output 3

6
6
6
6
6

Any two balls have equal numbers written on them.

Sample Input 4

8
1 2 1 4 2 1 4 1

Sample Output 4

5
7
5
7
7
5
7
5

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Globalization;
using System.Diagnostics;


class Solve
{
  public Solve() { }
  public static int Main()
  {
    new Solve().calc();
    return 0;
  }


  Scanner cin;

  void calc()
  {
    cin = new Scanner();
    int N = cin.nextInt();
    int[] all = new int[N];
    Dictionary<int, int> dic = new Dictionary<int, int>();
    for (int i = 0; i < N; i++)
    {
      int next = cin.nextInt();
      if (dic.ContainsKey(next))
      {
        dic[next]++;
      }
      else
      {
        dic[next] = 1;
      }
      all[i] = next;
    }

    int result = 0;
    foreach(var kvp in dic)
    {
      int n = dic[kvp.Key];
      int com = n >= 2 ? (n * (n - 1)) / 2 : 0;
      result += com;
    }

    for (int i=0; i<N; i++)
    {
      int num = all[i];
      int lineResult = result;
      if (dic[num] > 2)
      {
        lineResult -= dic[num] - 1;
      }
      else if (dic[num] == 2)
      {
        lineResult -= 1;
      }
      Console.WriteLine(lineResult.ToString());
    }
  }
}
class Modulo
{
  private const int M = 1000000007;
  private readonly int[] m_facs;
  public int Mul(int a, int b)
  {
    return (int)(Math.BigMul(a, b) % M);
  }
  public Modulo(int n)
  {
    m_facs = new int[n + 1];
    m_facs[0] = 1;
    for (int i = 1; i <= n; ++i)
    {
      m_facs[i] = Mul(m_facs[i - 1], i);
    }
  }
  public int Fac(int n)
  {
    return m_facs[n];
  }
  public int Pow(int a, int m)
  {
    switch (m)
    {
      case 0:
        return 1;
      case 1:
        return a;
      default:
        int p1 = Pow(a, m / 2);
        int p2 = Mul(p1, p1);
        return ((m % 2) == 0) ? p2 : Mul(p2, a);
    }
  }
  public int Div(int a, int b)
  {
    return Mul(a, Pow(b, M - 2));
  }
  public int Ncr(int n, int r)
  {
    if (n < r) { return 0; }
    if (n == r) { return 1; }
    int res = Fac(n);
    res = Div(res, Fac(r));
    res = Div(res, Fac(n - r));
    return res;
  }
}

class Scanner
{
  string[] s;
  int i;

  char[] cs = new char[] { ' ' };

  public Scanner()
  {
    s = new string[0];
    i = 0;
  }

  public string next()
  {
    if (i < s.Length) return s[i++];
    string st = Console.ReadLine();
    while (st == "") st = Console.ReadLine();
    s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
    if (s.Length == 0) return next();
    i = 0;
    return s[i++];
  }

  public int nextInt()
  {
    return int.Parse(next());
  }
  public int[] ArrayInt(int N, int add = 0)
  {
    int[] Array = new int[N];
    for (int i = 0; i < N; i++)
    {
      Array[i] = nextInt() + add;
    }
    return Array;
  }

  public long nextLong()
  {
    return long.Parse(next());
  }

  public ulong nextULong()
  {
    return ulong.Parse(next());
  }

  public long[] ArrayLong(int N, long add = 0)
  {
    long[] Array = new long[N];
    for (int i = 0; i < N; i++)
    {
      Array[i] = nextLong() + add;
    }
    return Array;
  }

  public double nextDouble()
  {
    return double.Parse(next());
  }


  public double[] ArrayDouble(int N, double add = 0)
  {
    double[] Array = new double[N];
    for (int i = 0; i < N; i++)
    {
      Array[i] = nextDouble() + add;
    }
    return Array;
  }
}