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 (GCC 5.4.1) (judge verdict: Wrong Answer)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define c_cin(X) scanf("%c",&X);
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  s_coutc(X) printf("%s",X)

#define  c_cout(X) printf("%c",X)


#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

//#define pai  3.1415926535      
#define deg_rad(X)  (X*(M_PI/180))

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

#pragma endregion

//137-C
static char  s[100001][11];
int main(void) {
	int   n, i, ans = 0;

	//各数文字を昇順 
	//それぞれの数字を昇順
	//同じ数字を探すmemcmp?
	i_cin(n);
	for (i = 0; i < n; i++) {
		char a[11];
		s_cin(s[i]); s[i][10] = 0x00;
		qsort(s[i], 10, sizeof(char), s_asorts);
	}

	qsort(s, n, 11, s_asorts);

	int cnt = 0;
	for (i = 0; i < n - 1; i++) {
		if (memcmp(s[i], s[i + 1], 10)) {
			if (cnt > 1) {
				ans += (cnt * cnt - 1) / 2;
			}
			cnt = 0;
			continue;
		}
		ans++; cnt++;
	}
	i_cout(ans);
	return 0;
}


//144-C
//

//157-C
//int main(void) {
//	int n, m;
//	int s, c;
//	char a[4] = { 0,0,0,0 };
//	int i, ans=0;
//	
//	i_cin2(n, m);
//
//	for (i = 0; i < m; i++) {
//		i_cin2(s, c);
//		if (n > 1 && s == 1 && !c) continue;
//		if(a[s-1])
//		  a[s - 1] = _min(a[s - 1], c + '0');
//		else 
//		  a[s - 1] = c + '0';
//	}
//	for (i = 0; i < n; i++)
//		if (!a[i])a[i] ='0';
//
//	a[n] = 0;
//	ans = atoi(a);
//	if (n == 3 && (ans < 10 || ans < 100)) ans = -1;
//	if (n == 2 && ans < 10) ans = -1;
//	i_cout(ans);
//	return 0;
//}

	
 //ABC-07 D 幅優先探索
//#define Q_SIZE 100000
//static struct _queue {
//	char *d[Q_SIZE];
//	int h;
//	int t;
//} q;
//
//int  pushq(struct _queue* q, char *item);
//int  pushq(struct _queue* q, char *item) {
//	if (q->t == Q_SIZE) return -1;
//	q->d[q->t] = item;
//	q->t++;
//	return 0;
//}
//
//char* popq(struct _queue* q);
//char* popq(struct _queue* q) {
//	char* tmp;
//	if (q->h == q->t) return -1;
//	tmp = q->d[q->h];
//	q->h--;
//	return tmp;
//}
//
//static struct _p {
//	int x;
//	int y;
//	int cnt;
//} q;
//static char a[51][51];
//static struct _p p[50];
//static int    pcnt;
//
//int _enq(struct _p *p, int x, int y, int cnt);
//int _enq(struct _p *p,int x, int y, int cnt) {
//	p->x = x; p->y = y; p->cnt = cnt;
//	pushq(&q,p); pcnt++;
//}
//int _deq(struct _p* p);
//int _deq(struct _p* p) {
//	p=popq(&q);
//	return(p == -1 ? p : 0);
//}
//int sqr_chk(int x, int y, int cnt) {
//	int tmp;
//	tmp = pcnt;
//	if (a[x - 1][y - 1] == '.') 
//	{
//		_enq(&p[pcnt], x - 1, y - 1, cnt); a[x - 1][y - 1] = 'X';
//	}
//	if (a[x - 1][y] == '.')
//	{
//		_enq(&p[pcnt], x - 1, y, cnt); a[x - 1][y] = 'X';
//	}
//	if (a[x+1][y] == '.') 
//	{
//		_enq(&p[pcnt], x + 1, y, cnt); a[x + 1][y] = 'X';
//	}
//	if (a[x][y+1] == '.') 
//	{
//		_enq(&p[pcnt], x, y + 1, cnt); a[x][y + 1] = 'X';
//	}
//	if (pcnt == tmp) return -1;
//	else             return 0;
//}
//
//int main(void) {
//	int r, c;
//	int sy, sx, gy, gx;
//	int i = 0, j = 0, x = 0, y = 0, ans = 0;
//	char cnt = '0';
//
//	i_cin(r, c);
//	i_cin4(sx, sy, gx, gy);
//	if (sx == gx && sy == gy) {
//		i_cout(0); return 0;
//	}
//	for (i = 0; i < r; i++)
//		s_cin(a[i]);
//	sy--; sx--;
//	struct _p* pt;
//	pt = &p[cnt];
//	p->x = sx; p->y = sy; p->cnt = cnt;
//
//	int rc;
//	while (1) {
//		rc = sqr_chk(pt->x,pt->y,pt->cnt);
//		if (rc == -1) {
//			rc = _deq(pt);
//			if (rc == -1 || (pt->x == gx && pt->y == gy)) break;
//		}
//	}
//
//	if (rc == -1) i_cout(rc);
//	else i_cout(ans);
//	return 0;
//}