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. 1 ≤ *i*, *j* ≤ *N*

2. *S*[*i*] = *S*[*j*], that is the *i*-th symbol of string *S* is equal to the *j*-th.
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 10<sup class="upper-index">5</sup>.
n=input()
count=0
d={}
for i in n:
    if i not in d:
        d[i]=1
    else:
        d[i]+=1
for i in d.values():
    count+=i**2
print(count)
'''
class Person:
    _num=0
        
    def __init__(self,name,sex,birthday,ident):
        if not (isinstance(name,str) and sex in ("女","男")):
            raise PersonValueError(name,sex)
        try:
            birth=datetime.date(*birthday) #生成一个日期对象
        except:
            raise PersionValueError("Wrong date:",birthday)
            self._name=name
            self._sex=sex
            self._birthday=birth
            self._id=ident
            Person._num+=1    #实例计数


    def id(self): return self._id
    def name(self): return self._name
    def sex(self): return self._sex
    def birthday(self): return self._birthday
    def age(self): return (datetime.date.today().year().year-self._birthday.year)

    def set_name(self,name):
        if not isinstance(name,str):
            raise PersonValueError("set_name",name)
    def __lt__(self,another):
        if not isinstance(another,Person):
            raise PersonTypeError(another)
        return self._id<another._id
'''