Problem: 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.

Input

The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.

Output

In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.

Examples

Input

aba


Output

First


Input

abca


Output

Second

Solution (in PYTHON3):

Solution:
from sys import stdin,stdout,setrecursionlimit
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush,nlargest
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm , accumulate
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
setrecursionlimit(10**9)
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
#def data(): return sys.stdin.readline().strip()
def data(): return input()
def num():return int(input())
def L(): return list(sp())
def LF(): return list(spf())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def spf(): return map(int, input.readline().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def pmat(A):
    for ele in A:
        print(*ele,end="\n")
def pmat2(A):
    for ele in A:
        for j in ele:
            print(j,end='')
        print()
def iseven(n):
    return n%2==0

def seive(r):
    prime=[1 for i in range(r+1)]
    prime[0]=0
    prime[1]=0
    for i in range(r+1):
        if(prime[i]):
            for j in range(2*i,r+1,i):
                prime[j]=0
    return prime
#solution
#ACPC 
#remeber cut ribbon problem
# set data structures faster than binary search sometimes
#bipartite match dfs
#think in problems with recursive manner.
a=input()
HM={}
for i in a:
    HM[i]=HM.get(i,0)+1
odd_count=0
for i,v in HM.items():
    if v%2==1:
        odd_count+=1
if odd_count<2:
    print("First")
else:
    if odd_count%2==0:
        print("Second")
    else:
        print("First")


    
endtime = time.time()
#print(f"Runtime of the program is {endtime - starttime}")