Problem:
Score : 500 points

Problem Statement

There is a string s of length 3 or greater.
No two neighboring characters in s are equal.

Takahashi and Aoki will play a game against each other.
The two players alternately performs the following operation, Takahashi going first:

Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.

The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Constraints

3 ≤ |s| ≤ 10^5

s consists of lowercase English letters.

No two neighboring characters in s are equal.

Input

The input is given from Standard Input in the following format:

s

Output

If Takahashi will win, print First. If Aoki will win, print Second.

Sample Input 1

aba

Sample Output 1

Second

Takahashi, who goes first, cannot perform the operation, since removal of the b, which is the only character not at either ends of s, would result in s becoming aa, with two as neighboring.

Sample Input 2

abc

Sample Output 2

First

When Takahashi removes b from s, it becomes ac.
Then, Aoki cannot perform the operation, since there is no character in s, excluding both ends.

Sample Input 3

abcab

Sample Output 3

First

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// tested by Hightail: https://github.com/dj3500/hightail
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <algorithm>
#include <sstream>
#include <stack>
#include <cstring>
#include <iomanip>
#include <ctime>
#include <cassert>
using namespace std;
#define pb push_back
#define INF 1001001001
#define FOR(i,n) for(int (i)=0;(i)<(n);++(i))
#define FORI(i,n) for(int (i)=1;(i)<=(n);++(i))
#define mp make_pair
#define pii pair<int,int>
#define ll long long
#define vi vector<int>
#define SZ(x) ((int)((x).size()))
#define fi first
#define se second
#define wez(n) int (n); scanf("%d",&(n));
#define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m));
#define wez3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k));
inline void pisz(int n) { printf("%d\n",n); }
template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";}
template<typename T> ostream& operator<<(ostream &s,vector<T> t){FOR(i,SZ(t))s<<t[i]<<" ";return s; }
#define DBG(vari) cout<<"["<<__LINE__<<"] "<<#vari<<" = "<<(vari)<<endl;
#define ALL(t) t.begin(),t.end()
#define FOREACH(i,t) for (__typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define TESTS wez(testow)while(testow--)
#define REP(i,a,b) for(int (i)=(a);(i)<=(b);++i)
#define REPD(i,a,b) for(int (i)=(a); (i)>=(b);--i)
#define REMAX(a,b) (a)=max((a),(b));
#define REMIN(a,b) (a)=min((a),(b));
#define IOS ios_base::sync_with_stdio(0);

int main () {
   string s;
   cin >> s;
   bool can = 0;
   REP(i,1,SZ(s)-2) if (s[i-1] != s[i+1]) can = 1;
   if (!can) cout << "Second"; else cout << "First";
}