output_description: If Takahashi will win, print `First`. If Aoki will win, print `Second`.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      s
attempt: #pragma GCC optimize ("O3")
  #pragma GCC target ("tune=native")
  #pragma GCC target ("avx")
  
  #include <bits/stdc++.h>
  
  // 汎用マクロ
  #define ALL_OF(x) (x).begin(), (x).end()
  #define REP(i,n) for (long long i=0, i##_len=(n); i<i##_len; i++)
  #define RANGE(i,is,ie) for (long long i=(is), i##_end=(ie); i<=i##_end; i++)
  #define DSRNG(i,is,ie) for (long long i=(is), i##_end=(ie); i>=i##_end; i--)
  #define UNIQUE(v) { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); }
  template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b; return true;} return false; }
  template<class T> bool chmin(T &a, const T &b) {if (a > b) {a = b; return true;} return false; }
  #define INF 0x7FFFFFFF
  #define LINF 0x7FFFFFFFFFFFFFFFLL
  #define Yes(q) ((q) ? "Yes" : "No")
  #define YES(q) ((q) ? "YES" : "NO")
  #define Possible(q) ((q) ? "Possible" : "Impossible")
  #define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")
  #define DUMP(q) cerr << "[DEBUG] " #q ": " << (q) << " at " __FILE__ ":" << __LINE__ << endl
  #define DUMPALL(q) { cerr << "[DEBUG] " #q ": ["; REP(i, (q).size()) { cerr << (q)[i] << (i == i_len-1 ? "" : ", "); } cerr << "] at " __FILE__ ":" << __LINE__ << endl; }
  template<class T> T gcd(const T &a, const T &b) { return a < b ? gcd(b, a) : b ? gcd(b, a % b) : a; }
  template<class T> T lcm(const T &a, const T &b) { return a / gcd(a, b) * b; }
  
  // gcc拡張マクロ
  #define popcount __builtin_popcount
  #define popcountll __builtin_popcountll
  
  // エイリアス
  #define DANCE_ long
  #define ROBOT_ unsigned
  #define HUMAN_ signed
  #define CHOKUDAI_ const
  using  ll = DANCE_ HUMAN_ DANCE_;
  using ull = DANCE_ ROBOT_ DANCE_;
  using cll = DANCE_ DANCE_ CHOKUDAI_;
  using  ld = long double;
  using namespace std;
  
  // モジュール
  
  
  // 処理内容
  int main() {
      
      string s; cin >> s;
      ll n = s.size();
      
      cout << ((s.front() == s.back()) != !!(n % 2) ? "First" : "Second") << endl;
      
  }
problem_description: 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.
samples: [{"input": "aba", "output": "Second\n    \n\nTakahashi, who goes first, cannot perform the operation, since removal of the\n`b`, which is the only character not at either ends of s, would result in s\nbecoming `aa`, with two `a`s neighboring.\n\n* * *"}, {"input": "abc", "output": "First\n    \n\nWhen Takahashi removes `b` from s, it becomes `ac`. Then, Aoki cannot perform\nthe operation, since there is no character in s, excluding both ends.\n\n* * *"}, {"input": "abcab", "output": "First"}]