After learning a lot about space exploration, a little girl named Ana wants to change the subject.

Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:

You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa"). 

Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).

Input

The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.

Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array. 

The total number of characters in the input array will be less than 1 000 000.

Output

Output one number, representing how many palindrome pairs there are in the array.

Examples

Input

3
aa
bb
cd


Output

1


Input

6
aab
abcac
dffe
ed
aa
aade


Output

6

Note

The first example:

  1. aa + bb → abba. 



The second example:

  1. aab + abcac = aababcac → aabccbaa
  2. aab + aa = aabaa
  3. abcac + aa = abcacaa → aacbcaa
  4. dffe + ed = dffeed → fdeedf
  5. dffe + aade = dffeaade → adfaafde
  6. ed + aade = edaade → aeddea
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx4[] = {0, 0, +1, -1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dy8[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
struct edge {
  int u, v, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  string s;
  map<long long, long long> mp;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> s;
    long long a[30];
    long long b[30];
    memset(a, 0, sizeof(a));
    int l = s.length();
    for (int j = 0; j < l; j++) a[s[j] - 'a']++;
    for (int j = 0; j < 26; j++) {
      a[j] %= 2;
      b[j] = a[j];
    }
    long long num = 0, p = 1;
    for (int j = 25; j >= 0; j--) {
      num += (a[j] * p);
      p *= 2;
    }
    p = 1;
    for (int j = 0; j < 26; j++) {
      b[j] = 1 - b[j];
      long long num1 = 0;
      p = 1;
      for (int j = 25; j >= 0; j--) {
        num1 += (b[j] * p);
        p *= 2;
      }
      ans += mp[num1];
      b[j] = 1 - b[j];
    }
    ans += mp[num];
    mp[num]++;
  }
  cout << ans << '\n';
}
