Problem:
Score : 300 points

Problem Statement

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.

For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.

Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Constraints

2 \leq N \leq 10^5

s_i is a string of length 10.

Each character in s_i is a lowercase English letter.

s_1, s_2, \ldots, s_N are all distinct.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Sample Input 1

3
acornistnt
peanutbomb
constraint

Sample Output 1

1

s_1 = acornistnt is an anagram of s_3 = constraint. There are no other pairs i, j such that s_i is an anagram of s_j, so the answer is 1.

Sample Input 2

2
oneplustwo
ninemodsix

Sample Output 2

0

If there is no pair i, j such that s_i is an anagram of s_j, print 0.

Sample Input 3

5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

Sample Output 3

4

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)

class Combination:                                      # 計算量は O(n_max + log(mod))
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        f = 1
        self.fac = fac = [f]
        for i in range(1, n_max+1):                     # 階乗(= n_max !)の逆元を生成
            f = f * i % mod                             # 動的計画法による階乗の高速計算
            fac.append(f)                               # fac は階乗のリスト
        f = pow(f, mod-2, mod)                          # 階乗から階乗の逆元を計算。フェルマーの小定理より、　a^-1 = a^(p-2) (mod p) if p = prime number and p and a are coprime
                                                        # python の pow 関数は自動的に mod の下での高速累乗を行ってくれる
        self.facinv = facinv = [f]
        for i in range(n_max, 0, -1):                   # 上記の階乗の逆元から階乗の逆元のリストを生成（＝ facinv ）
            f = f * i % mod
            facinv.append(f)
        facinv.reverse()


    def __call__(self, n, r):  # self.C と同じ
        if not 0 <= r <= n: return 0
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def C(self, n, r):
        if not 0 <= r <= n: return 0                    # 範囲外という事はすなわち、そのような場合の数は無いはずなので、0を出力すればよい。（これが無いとREになる）
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

##############################################################################################

from collections import Counter
N = int(input())
S = [[0]*26 for _ in range(N)]
for i in range(N):
    for s in input():
        S[i][ord(s)-ord("a")] += 1
    S[i] = tuple(S[i])
C = Combination(N)
res = 0
for value in Counter(S).values():
    res += C(value,2)

print(res)