Choosing Symbol Pairs

Description

There is a given string S consisting of N symbols. Your task is to find the number of ordered pairs of integers i and j such that

1 ≤ i, j ≤ N

S[i] = S[j], that is the i-th symbol of string S is equal to the j-th.

Input

The single input line contains S, consisting of lowercase Latin letters and digits. It is guaranteed that string S in not empty and its length does not exceed 105.

Output

Print a single number which represents the number of pairs i and j with the needed property. Pairs (x, y) and (y, x) should be considered different, i.e. the ordered pairs count.

Sample Input

Input

great10

Output

7

Input

aaaaaaaaaa

Output

100

题目很水，就是让你配对，题意也很好理解，注意int不行，会越界,我错了两次就是没有进行好强制转换，看一下我的代码吧

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <algorithm>
using namespace std;
int main()
{
char x[100005];
int a[50];
while(~scanf("%s", x))
{
memset(a, 0, sizeof(a));
int len = strlen(x);
for(int i = 0; i < len; i++)
{
if(islower(x[i])) a[x[i] - 'a' + 10]++;
else a[x[i] - '0']++;
}
long long sum = 0;
for(int i = 0; i < 36; i++)
{
sum += (a[i] * 1LL *a[i]);
}
printf("%lld\n", sum);
}
return 0;
}