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

////////////////////////////////////////
///  tu3 pro-con template            ///
////////////////////////////////////////
#include "bits/stdc++.h"
using namespace std;

// -- typedefs -- //
#define EPS 1e-9
typedef long long llong;

// -- loop macros -- //
#define REP(i,n) for (auto i = decltype(n)(0); i < (n); i++)
#define RREP(i,n) for (auto i = (n)-1; i >= 0; i--)
#define FOR(i,s,n) for (auto i = (s); i < (n); i++)
#define FORR(i,s,n) for (auto i = (n)-1; i >= (s); i--)
#define FORE(exp) for (auto && exp)
#define allof(c) c.begin(), c.end()
#define partof(c,s,e) c.begin() + (s), c.begin() + (e)

// -- functors -- //
#define PREDICATE(t,a,...) [&](const t & a) -> bool { return __VA_ARGS__; }
#define PRED(a,...) PREDICATE(auto,a,__VA_ARGS__)
#define COMPARISON(t,a,b,...) [&](const t & a, const t & b) -> bool { return __VA_ARGS__; }
#define COMP(a,b,...) COMPARISON(auto,a,b,__VA_ARGS__)
#define CONV1(a,...) [&](const auto & a) -> auto { return __VA_ARGS__; }
#define CONV2(a,b,...) [&](const auto & a, const auto & b) -> auto { return __VA_ARGS__; }
#define CONV3(a,b,c,...) [&](const auto & a, const auto & b, const auto & c) -> auto { return __VA_ARGS__; }

// -- I/O Helper -- //
struct _Reader { istream& cin; template <class T> _Reader operator ,(T& rhs) { cin >> rhs; return *this; } };
struct _Writer { ostream& cout; const char* d{ " " }; bool f{}; template <class T> _Writer operator ,(const T& rhs) { cout << (f ? d : "") << rhs; f = 1; return *this; } };
#define READ(t,...) t __VA_ARGS__; (_Reader{cin}), __VA_ARGS__
#define WRITE(...) do { (_Writer{cout}), __VA_ARGS__; cout << endl; } while (false)
#define WRITELN(...)  do { (_Writer{cout, "\n"}), __VA_ARGS__; cout << endl; } while (false)
#ifdef _DEBUG
#define DEBUG(...) (_Writer{cerr}), "DEBUG -> ", __VA_ARGS__, "\n"
#else
#define DEBUG(...)
#endif

// -- vevector -- //
template <class T> struct vevector : vector<vector<T>> { vevector(size_t n = 0, size_t m = 0, const T & initial = T()) : vector<vector<T>>(n, vector<T>(m, initial)) { } };
template <class T> struct vevevector : vector<vevector<T>> { vevevector(size_t n = 0, size_t m = 0, size_t l = 0, const T & initial = T()) : vector<vevector<T>>(n, vevector<T>(m, l, initial)) { } };
template <class T> struct vevevevector : vector<vevevector<T>> { vevevevector(size_t n = 0, size_t m = 0, size_t l = 0, size_t k = 0, const T & initial = T()) : vector<vevevector<T>>(n, vevevector<T>(m, l, k, initial)) { } };

namespace std {
	template <class T1, class T2> istream& operator >> (istream& in, pair<T1, T2>& p) { in >> p.first >> p.second; return in; }
	template <class T1, class T2> ostream& operator << (ostream& out, const pair<T1, T2>& p) { out << p.first << " " << p.second; return out; }
}

template <class T> T read() { T t; cin >> t; return t; }
template <class T> vector<T> read(int n) { vector<T> v; REP(i, n) { v.push_back(read<T>()); } return v; }
template <class T> vevector<T> read(int n, int m) { vevector<T> v; REP(i, n) v.push_back(read<T>(m)); return v; }
template <class T> vector<T> readjag() { return read<T>(read<int>()); }
template <class T> vevector<T> readjag(int n) { vevector<T> v; REP(i, n) v.push_back(readjag<T>()); return v; }

template <class T> struct iter_range_t { T beg, end; };
template <class T> iter_range_t<T> iter_range(T beg, T end) { return iter_range_t<T>{beg, end}; }
template <class T> ostream& operator << (ostream& out, iter_range_t<T> v) { if (v.beg != v.end) { out << *v.beg++; while (v.beg != v.end) { out << " " << *v.beg++; } } return out; }
template <class T> ostream& operator << (ostream& out, const vector<T>& v) { return out << iter_range(allof(v)); }

// -- etc -- //
template <class T> T infinity_value();
template <> int infinity_value<int>() { return int(1) << 28; }
template <> llong infinity_value<llong>() { return llong(1) << 60; }
template <> double infinity_value<double>() { return 1e+300 * 1e+300; }
#define INF(T) infinity_value<T>()

inline int sign_of(double x) { return (abs(x) < EPS ? 0 : x > 0 ? 1 : -1); }
template <class TInt> bool in_range(TInt val, TInt min, TInt max) { return val >= min && val < max; }
template <> bool in_range<double>(double val, double min, double max) { return val - min > -EPS && val - max < EPS; }
template <class TInt> bool in_range2d(TInt x, TInt y, TInt w, TInt h) { return x >= 0 && x < w && y >= 0 && y < h; }
vector<int> iotavn(int start, int count) { vector<int> r(count); iota(allof(r), start);	return r; }


//// start up ////
void solve();
int32_t main()
{
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed;
	cout << setprecision(std::numeric_limits<double>::max_digits10);
	solve();

	return 0;
}

////////////////////
//// template end
////////////////////
void solve()
{
	READ(int, N);
	map<string, llong> input;
	REP(i, N)
	{
		READ(string, s);
		sort(allof(s));
		input[s]++;
	}

	llong ans = 0;
	FORE(p: input)
	{
		ans += (p.second) * (p.second - 1) / 2;
	}
	WRITE(ans);
}