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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

#define lli long long
#define all(i) i.begin(),i.end()
#define rall(i) i.rbegin(),i.rend()
#define rep0(i,j) for(int i=0;i<j;i++)
#define rep1(i,j) for(int i=1;i<=j;i++)
#define rep0d(i,j) for(int i=j-1;i>=0;i--)
#define MOD 1000000007
#define MAX (lli)1e12 
typedef pair<lli, lli> p; 
class Graph{
  int V;
  vector<list<p>> adj; 
  const int INF = (int)1e6;

  public: 
  Graph(lli V): adj(V){ 
    this->V = V; 
  } 

  void addEdge(int u, int v, int w){ 
    adj[u].push_back(make_pair(v, w)); 
  } 

  lli edgeNum(){
    lli sum=0;
    rep0(i,V)sum+=adj[i].size();
    return sum;
  }

  bool route(lli from, lli to){//vec[i]==j: j->iっていうもとに戻れる道がある
    vector<lli> vec(V, -1);
    dfs(1,from,vec);
    return vec[to] >= 0;
  }

  void dfs(lli ed, lli num, vector<lli>& vec){
    for(auto e: adj[num]){
      if(vec[e.first] != -1)continue;
      vec[e.first] = num;
      dfs(ed+1, e.first, vec);
    }
  }

  vector<int> bfs(int n){
    queue<int> q;
    vector<int> vec(V,-1);

    q.push(n);
    vec[n]=0;
    while(!q.empty()){
      int p=q.front();
      q.pop();
      for(auto e:adj[p]){
        if(vec[e.first]==-1){
          q.push(e.first);
          vec[e.first] = e.second + vec[p];
        }
      }
    }
    return vec;
  }

  vector<int> shortestPath(int src){ 
    priority_queue< p, vector <p> , greater<p> > pq; 
  
    vector<int> dist(V, INF); 
    pq.push(make_pair(0, src)); 
    dist[src] = 0; 
  
    while (!pq.empty()) { 
        int u = pq.top().second; 
        pq.pop(); 
  
        list<p>::iterator i; 
        for (auto i = adj[u].begin(); i != adj[u].end(); ++i) { 
            int v = i->first; 
            int weight = i->second; 
  
            if (dist[v] > dist[u] + weight) { 
                dist[v] = dist[u] + weight; 
                pq.push(make_pair(dist[v], v)); 
            } 
        } 
    } 
    for(int i=0;i<V;i++)if(dist[i]==INF)dist[i]=-1;
    return dist;
  } 

};

struct UF{
    vector<int> par,size;

    UF(int N) : par(N), size(N, 1){
        rep0(i,N)par[i] = i;
    }

    int root(int x){
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y){
        int rx = root(x);
        int ry = root(y);
        if(rx == ry) return;
        par[rx] = ry;
        size[rx] += size[ry];
    }

    bool same(int x, int y){
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    int getSize(int x){
        return size[x];
    }
};
struct BIT{
  vector<lli> tab;//bit table
  BIT(int n):tab(n+1){
    rep0(i,n+1)tab[i]=0;
  }

  void add(int p, lli num){
    for(++p;p<tab.size();p+=p&-p)tab[p]+=num;
  }

  lli sum(int p){//from 0 to p
    lli ans=0;
    for(++p;p>0;p-=p&-p)ans+=tab[p];
    return ans;
  }

  void debug(){
    rep0(i,tab.size())cout << tab[i] << ' ';
    cout << endl;
  }
};
 
//pairに関するもの
lli sum(p c){
  return (c.first + c.second);
}
p operator-(const p & l,const p & r) {   
    return {l.first-r.first,l.second-r.second};                                    
}
bool cmpf(const pair<lli, lli> left, const pair<lli, lli> right) {
    return left.first < right.first;
}
bool cmps(const pair<lli, lli> left, const pair<lli, lli> right) {
    return left.second < right.second;
}
 
lli gcd(lli a, lli b){
    if(a < b)return gcd(b, a);
    lli temp;
    while((temp = a%b)){
        a=b;
        b=temp;
    }
    return b;
}

struct nCr{
  vector<lli> f, rf;
  lli count=0;
  nCr(lli num): f(num), rf(num){
      count = num;
      f[0] = 1;
      rep1(i, num) f[i] = (f[i - 1] * i) % MOD;
      rep0(i, num+1) rf[i] = inv(f[i]);
  }
    
  lli inv(lli x) {
      lli res = 1;
      lli k = MOD - 2;
      lli y = x;
      while (k) {
          if (k & 1) res = (res * y) % MOD;
          y = (y * y) % MOD;
          k /= 2;
      }
      return res;
  }
  lli C(lli n, lli k) {
      if(k > n || k<0)return 0;
      lli a = f[n];
      lli b = rf[n-k];
      lli c = rf[k];
   
      lli bc = (b * c) % MOD;
      return (a * bc) % MOD;
  }
};
int main(){
  lli n;
  cin >> n;
  string str[n];
  vector<vector<char>> st(n,vector<char>(10));
  rep0(i,n){
    cin >> str[i];
    rep0(j,10)st[i][j]=str[i][j];
    sort(all(st[i]));
  }
  lli ans=0;
  rep0(i,n)rep0(j,i){
    ans+=st[i]==st[j];
  }
  cout << ans;
}