The Little Girl loves problems on games very much. Here's one of them.

Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
 -  The players move in turns; In one move the player can remove an arbitrary letter from string *s*. -  If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't. 
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
The input contains a single line, containing string *s* (1 ≤ |*s*|  ≤  10<sup class="upper-index">3</sup>). String *s* consists of lowercase English letters.


"""
rules:
player can remove arbitrary letter from s
if the player can reorder letter from s to get a palindrome this player wins


palindromes:

abca
First: abba
Second: won

abab



aabbaa
ababaa


aba
aab

abcba
aabbc

update condition
  remove a letter with an odd count

reorder condition:
if len is even
  check if counts of every str in even
if len is odd:
  the num of odd count letters should be 1


aaa
"""
counts = {}
s = list(input())
for c in s:
  if c not in counts: counts[c] = 1
  else: counts[c] += 1
odd_lets = []
for k in counts:
  if counts[k] % 2 != 0: odd_lets.append(k)
is_even = len(s) % 2 == 0
r=False
while (is_even and len(odd_lets) > 0) or (not is_even and len(odd_lets) > 1):
  counts[odd_lets[-1]] -=1
  is_even = not is_even
  print(odd_lets[-1], counts[odd_lets[-1]])
  if counts[odd_lets[-1]] == 0:
    counts.pop(odd_lets[-1])
    odd_lets.pop()
  r = not r
print('First' if r == False else 'Second')
  