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>.
import time

def feibo(n):            

       f1=1

       f2=1

       if n%2!=0:

              for i in range(1,n//2+2):

                     f1=f1+f2

                     f2=f1+f2

              return f1

       elif n%2==0:

              for i in range(1,n//2+1):

                     f1=f1+f2

                     f2=f1+f2

              return f2

n=eval(input("请输入月份"))

s=time.time()

feibo(n)

e=time.time()

print(e-s)