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++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const int CM = 130000;
char cn[CM + 10], * ci = cn + 10, * owa = cn + CM;
const ll ma0 = 1157442765409226768;
const ll ma1 = 1085102592571150095;
const ll ma2 = 71777214294589695;
const ll ma3 = 281470681808895;
const ll ma4 = 4294967295;
inline int getint() {
	char c = ci[8];
	ci[8] = 0;
	ll tmp = *(ll*)ci;
	int dig = 68 - __builtin_ctzll((tmp & ma0) ^ ma0);
	tmp = tmp << dig & ma1;
	tmp = tmp * 10 + (tmp >> 8) & ma2;
	tmp = tmp * 100 + (tmp >> 16) & ma3;
	tmp = tmp * 10000 + (tmp >> 32) & ma4;
	ci[8] = c;
	ci += (64 - dig >> 3) + 1;

	if (ci - owa > 0) {
		memcpy(cn, owa, 10);
		ci -= CM;
		fread(cn + 10, 1, CM, stdin);
	}
	return tmp;
}

char* di = cn;
const int MAX = 200000;
class shuturyoku_unko {
public:
	char C[MAX * 6 - 100000];
	int K[MAX + 10];
	constexpr shuturyoku_unko() : C(), K() {
		auto di = C + 1;
		C[0] = '0';
		K[1] = 1;
		rep1(i, MAX) {
			int X = i;
			int keta = 0;
			char c[10] = {};
			while (X) {
				*(c + keta) = '0' + X % 10;
				X /= 10;
				keta++;
			}
			for (int j = keta - 1; j >= 0; j--)* di++ = (*(c + j));
			K[i + 1] = K[i] + keta;
		}
	}
} f;
inline void putint(ll X) {
	int ue = X / 100000;
	if (ue) {
		memcpy(di, f.C + f.K[ue], f.K[ue + 1] - f.K[ue]);
		di += f.K[ue + 1] - f.K[ue];
		int sita = X - ue * 100000ll;
		rep(i, 5 - f.K[sita + 1] + f.K[sita])* di++ = '0';
		memcpy(di, f.C + f.K[sita], f.K[sita + 1] - f.K[sita]);
		di += f.K[sita + 1] - f.K[sita];
	}
	else for (int i = f.K[X]; i < f.K[X + 1]; i++)* di++ = f.C[i];
	*di++ = '\n';
}

int B[200001];

int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);


	fread(ci, 1, CM, stdin);

	int N = getint();
	ll kotae = 1;
	int A[200000];
	rep(i, N) kotae += B[(A[i] = getint())]++;
	int *a = A;
	
	int amari = N % 10800;
	int shou = N / 10800;
	rep(j, shou) {
		rep(k, 10800) putint(kotae - B[*a++]);
		fwrite(cn, 1, di - cn, stdout);
		di = cn;
	}
	rep(k, amari) putint(kotae - B[*a++]);
	fwrite(cn, 1, di - cn, stdout);

	Would you please return 0;
}