Bubble Cup 11 - Finals [Online Mirror, Div. 2] H題

                                   H. Palindrome Pairs

                                                                       time limit per test  :2 seconds

                                                                     memory limit per test:256 megabytes

                                                                        input : standard input

output : standard output

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)(i,j) is considered the same as the pair (j,i)(j,i).

Input

The first line contains a positive integer N (1≤N≤100000), 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 1000000.

Output

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

Examples

input

Copy

​
3
aa
bb
cd

​
           

output

Copy

1
      

input

Copy

6
aab
abcac
dffe
ed
aa
aade
           

output

Copy

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

對于這個題，剛開始完全是使用暴力，但是T了兩發，很難受，靜下心 仔細一想，他所給的n是10^5,如果所有按照所有情況都給周遊一邊的話，那就得大約10^10了，必定T。

是以對于這個題的話，當組合成的串中，判斷字元相同的個數為單數時為1,個數為偶數或者沒有時為0,可以将字元串看成26位的01字元串，判斷字元串出現的次數。并且統計次數。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
string str;
map<int ,int>mp;
int n;
ll ans;
int main(){
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 1;i <= n;i++){
        int sum  = 0;
        cin >> str;
        for(int  j = 0;j < str.size();j++){
            int k = str[j] - 'a';
            sum ^= (1 << k);
        }
        ans += mp[sum];   
        for(int j = 0;j < 26;j++){
            ans += mp[sum ^ (1 << j)];
            if(ans) cout<<(sum ^ (1 << j))<<" ";
        }
        mp[sum]++;
    }
    cout <<ans<<endl;
    return 0;
}