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)

//AtCoder用
//プリプロセッサ系
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>
#define MOD 1000000007
#define INFTY 2147483646
#define DEBUG 1
#define END printf("\n");return 0;
#define QS09(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
#define QS90(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
//↓この2つはCE対策のために適当にdefineしてるだけだから使う際は「構造体クイックソート設定」をコピペ
#define ST_NAME _GD
#define KEY_NAME how_vertex

#define ST09(how_data,data) qsort(data,how_data,sizeof(ST_NAME),(int (*)(const void *,const void *))qsort_st09);
#define ST90(how_data,data) qsort(data,how_data,sizeof(ST_NAME),(int (*)(const void *,const void *))qsort_st90);
#define In(a) long a;scanf("%ld",&a);
#define Out(a) printf("%ld\n",a);
#define dprintf if(DEBUG)printf
#define ifdebug if(DEBUG)
#define MAKE_ARRAY_DATA In(how_data);long data[how_data];input_array(how_data,data);
#define update_max(a,b) a=(a>b?a:b);
#define update_min(a,b) a=(a<b?a:b);
#define input_long(A) int A;scanf("%d",&A)

//構造体クイックソート設定 使用時はmainに以下をコピペ、ST_NAMEとKEY_NAMEを書き換え
/*
    #undef ST_NAME
    #undef KEY_NAME
    #define ST_NAME _GD
    #define KEY_NAME how_vertex
    ST09(how_data,data);
*/


//関数
void input_array(long how_data,long *data); //配列の入力
void output_array(long how_data,long *data); //配列の出力
void input_array2(long first , long second , long data[][3]); //2次元配列の入力
void format_array(long how_data ,long *data,long what); //配列初期化
long get_random(long min, long max); //min～max範囲の乱数生成
long factorial(long n); //n!
long fibonacci(long n); //n番目のフィボナッチ
int qsort_09(const int *sys1 , const int *sys2);
int qsort_90(const int *sys1 , const int *sys2);
int qsort_st09(const void *c1, const void *c2);
int qsort_st90(const void *c1, const void *c2);
long sel_max(long a , long b);
long sel_min(long a , long b);
long array_max(long how_data,long *data);
long array_min(long how_data,long *data);
long array_search(long how_data,long *data,long what);
long can_DP(long how_data,long *data,long how_can,bool *can);
long array_sum(long how_data,long *data);
long Leven_dist(char *now , char *target);
long get_digit(long target);
long rounding(double target);
long roundingdown(double target);
long roundingup(double target);
long ncr(long n , long r);
long npr(long n , long r);
long nhr(long n , long r);
int fifw(long add,long *data);
long fifr(long *data);
void GRAPH_input(void);
void GRAPH_dfs(long now);
bool check_prime(long target);
long get_lcm(long a,long b);
long get_gcd(long a,long b);
void digit_cut(int how_data,char *s,int *i);  //文字列として与えられた数列を1桁ずつ分割
//グローバル変数一覧
long fifo[1000]={0};//fifo[0]はデータ件数
long loop1,loop2,loop3,loop4,loop5;
long g_ans=0; //answerグローバル変数用

typedef struct{
    long how_vertex;
    long how_edge;
    long edge_data[1000][2];
    bool aleady_searched_vertex[500];
    bool all_searched;
} _GD;//GraphData

_GD GRAPH={0,0,{{0}},{0},0};

char c_temp;
//loop5 loop4 loop3 loop2 loop1 input_long
//--------------ここから

int main(void){
    //全ての単語についてa～zの数の一致を見る
    //カウントO(N)+比較O(N^2) TLE
    //アルファベットに数字を割り振ってその和を比較(ソートできるのでO(log N))
    //虚解法
    In(how_data);
    char s[how_data][11];
    for(loop1=0;loop1<how_data;loop1++){
        scanf("%s",s[loop1]);
    }
    long s_evaluation[how_data];
    format_array(how_data,s_evaluation,0);
    //無理やり通すために誤った重複を排除したい→2乗すれば・・・？
    for(loop1=0;loop1<how_data;loop1++){
        for(loop2=0;loop2<10;loop2++){
            s_evaluation[loop1]+=((1+s[loop1][loop2]-'a')*(2+s[loop1][loop2]-'a'));
        }
        //printf("%ld\n",s_evaluation[loop1]);
    }
    QS09(how_data,s_evaluation);
    long ans=0;
    long ren=0;
    for(loop1=0;loop1<how_data-1;loop1+=ren){
        ren=0;
        for(loop2=loop1;loop2<how_data;loop2++){
            if(s_evaluation[loop1]==s_evaluation[loop2]){
                ren++;
            }else{
                break;
            }
        }
        //nCr=n!/r!(n-r)!
        ans+=factorial(ren)/2*factorial(ren-2);
    }
    printf("%ld",ans);
    END;
    
    
}






//--------------ここまで
 
void input_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        scanf("%ld",&data[loop]);
    }
    return ;
}
void output_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        printf("%ld ",data[loop]);
    }
    printf("\n");
    return ;
}
void input_array2(long first,long second,long data[][3]){
    long loopA,loopB;
    for(loopA=0;loopA<first;loopA++){
        for(loopB=0;loopB<second;loopB++){
            scanf("%ld",&data[loopA][loopB]);
        }
    }
    return ;
}
void format_array(long how_data ,long *data,long what){
    long loopA;
    for(loopA=0;loopA<how_data;loopA++){
        data[loopA]=what;
    }
    return ;
}
long get_random(long min, long max){   //指定の範囲から乱数を1つ返すやつ
    //srand((unsigned int)time(0));   //現在時刻で疑似乱数初期化
    rand();rand();rand();rand();    //乱数を空打ち
    return (rand()%(max+1-min))+min;
}
long factorial(long n){//n!のMOD10^9+7を返すやつ
    unsigned long long int ret=1;
    for(long i=1;i<=n;i++)ret=(ret*i)%1000000007;
    return (long)ret;
}
int qsort_09(const int *sys1 , const int *sys2){ //小さいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
    if(*sys1<*sys2){
        return -1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return 1;
    }
}
int qsort_90(const int *sys1, const int *sys2){ //大きいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
    if(*sys1<*sys2){
        return 1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return -1;
    }
}
int qsort_st09(const void *c1, const void *c2){
    ST_NAME hikaku1= *(ST_NAME *)c1;
    ST_NAME hikaku2= *(ST_NAME *)c2;
    return hikaku1.KEY_NAME - hikaku2.KEY_NAME;
}
int qsort_st90(const void *c1, const void *c2){
    ST_NAME hikaku1= *(ST_NAME *)c1;
    ST_NAME hikaku2= *(ST_NAME *)c2;
    return hikaku2.KEY_NAME - hikaku1.KEY_NAME;
}
long fibonacci(long n){
    switch(n){
        case 0:return 0;
        case 1:return 1;
        default :return fibonacci(n-1)+fibonacci(n-2);
    }
}
long sel_max(long a,long b){
    if(a>b)return a;
    return b;
}
long sel_min(long a,long b){
    if(a>b)return b;
    return a;
}

long can_DP(long how_data,long *data,long how_can,bool *can){//Typical DP Contest A
    //data内で組み合わせられる和をcanに0 or 1で入れる
    //返り値はパターン数
    long loopA,loopB;
    long ret=0;
    for(loopA=0;loopA<how_can;loopA++){//初期化
        can[loopA]=0;
    }
    can[0]=1;//絶対にありえる
    for(loopA=0;loopA<how_data;loopA++){
        for(loopB=how_can-1;loopB>=0;loopB--){
            if(can[loopB]==1 && loopB+data[loopA]<how_can){
                can[loopB+data[loopA]]=1;
            }
        }
    }
    for(loopA=0;loopA<how_can;loopA++){
        if(can[loopA]==1){
            ret++;
        }
    }
    return ret;
}
long array_max(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret<data[loop])ret=data[loop];
    }
    return ret;
}
long array_min(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret>data[loop])ret=data[loop];
    }
    return ret;
}
long array_sum(long how_data,long *data){
    long ret=0;
    long loop;
    for(loop=0;loop<how_data;loop++){
        ret+=data[loop];
    }
    return ret;
}
long array_search(long how_data,long *data,long what){
    long loop;
    for(loop=0;loop<how_data;loop++){
        if(data[loop]==what){
            return loop;
        }
    }
    return -1;
}
long Leven_dist(char *now , char *target){
    long loopA,loopB;
    //レーベンシュタイン距離を求める
    // 参考文献
    // http://nw.tsuda.ac.jp/class/algoB/c13.html  (アルゴリズム理解)
    // http://d.hatena.ne.jp/ohnishiakira/20090809/1249845529  (実装)
    long len_now=strlen(now)+1;
    long len_target=strlen(target)+1;
    long d[len_now][len_target]; //計算用
    for(loopA=0;loopA<len_now;loopA++) d[loopA][0]=loopA;
    for(loopA=0;loopA<len_target;loopA++) d[0][loopA]=loopA;
    for(loopA=1;loopA<len_now;loopA++){
        for(loopB=1;loopB<len_target;loopB++){
            long cost=(now[loopA-1]==target[loopB-1] ? 0:1);
            d[loopA][loopB]=sel_min(sel_min(d[loopA-1][loopB]+1,d[loopA][loopB-1]+1),d[loopA-1][loopB-1]+cost);
        }
    }
    return d[len_now-1][len_target-1];
}
long get_digit(long target){
    return (long)(log10(target)+1);
}
long ncr(long n , long r){
    //パスカルの三角形
    long loopA,loopB;
    long pascal[100][102]={{0}};
    pascal[1][0]=1;
    pascal[1][1]=1;
    for(loopA=2;loopA<100;loopA++){
        pascal[loopA][0]=1;
        for(loopB=1;loopB<loopA;loopB++){
            pascal[loopA][loopB]=pascal[loopA-1][loopB-1]+pascal[loopA-1][loopB];
        }
        pascal[loopA][loopA]=1;
    }
    return pascal[n][r];
}
long npr(long n, long r){
    return ncr(n,r)*factorial(r);
}
long nhr(long n , long r){
    return ncr(n+r-1,r);
}
long rounding(double target){
    return (long)target+0.50;
}
long roundingup(double target){
    return (long)ceil(target);
}
long roundingdown(double target){
    return (long)floor(target);
}
int fifw(long add,long *data){
    data[data[0]+1]=add;
    data[0]++;
    return data[0]; //現在のデータ件数を返す
}
long fifr(long *data){
    long ret=data[1];
    data[0]--;
    for(int i=1;i<999;i++){
        data[i]=data[i+1];
    }
    return ret;
}
void GRAPH_input(void){
    scanf("%ld %ld\n",&GRAPH.how_vertex,&GRAPH.how_edge);
    for(long loopA=0;loopA<GRAPH.how_edge;loopA++){
        scanf("%ld %ld\n",&GRAPH.edge_data[loopA][0],&GRAPH.edge_data[loopA][1]);
    }
    return;
}
void GRAPH_dfs(long now){
    GRAPH.aleady_searched_vertex[now]=1;
    for(int loopA=0;loopA<GRAPH.how_edge;loopA++){
        if(GRAPH.edge_data[loopA][0]==now && GRAPH.aleady_searched_vertex[GRAPH.edge_data[loopA][1]]==0){
            GRAPH_dfs(GRAPH.edge_data[loopA][1]);
        }
        if(GRAPH.edge_data[loopA][1]==now && GRAPH.aleady_searched_vertex[GRAPH.edge_data[loopA][0]]==0){
            GRAPH_dfs(GRAPH.edge_data[loopA][0]);
        }
    }
    return ;
}
bool check_prime(long target){//素数判定、素数なら1
    if(target==2)return 1;
    if(target<=1 || target%2==0)return 0;
    for(int loopA=3;loopA<=sqrt(target)+1;loopA+=2)if(target%loopA==0)return 0;
    return 1;
}
long get_lcm(long a,long b){//最小公倍数
    return (a*b)/get_gcd(a,b);
}
long get_gcd(long a,long b){//最大公約数 参考：https://webkaru.net/clang/find-gcd-of-two-number/
    // a>b
    if(a<b){
        int tem=a;
        a=b;
        b=tem;
    }
    long r=a%b;
    while(r!=0){
        a=b;
        b=r;
        r=a%b;
    }
    return b;
}
void digit_cut(int how_data,char *s,int *i){
    for(int loopA=0;loopA<how_data;loopA++){
        i[loopA]=atoi(&s[loopA]);
    }
    return ;
}