Problem:
Score : 200 points

Problem Statement

We will call a string that can be obtained by concatenating two equal strings an even string.
For example, xyzxyz and aaaaaa are even, while ababab and xyzxy are not.

You are given an even string S consisting of lowercase English letters.
Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S.
It is guaranteed that such a non-empty string exists for a given input.

Constraints

2 \leq |S| \leq 200

S is an even string consisting of lowercase English letters.

There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest even string that can be obtained.

Sample Input 1

abaababaab

Sample Output 1

6

abaababaab itself is even, but we need to delete at least one character.

abaababaa is not even.

abaababa is not even.

abaabab is not even.

abaaba is even. Thus, we should print its length, 6.

Sample Input 2

xxxx

Sample Output 2

2

xxx is not even.

xx is even.

Sample Input 3

abcabcabcabc

Sample Output 3

6

The longest even string that can be obtained is abcabc, whose length is 6.

Sample Input 4

akasakaakasakasakaakas

Sample Output 4

14

The longest even string that can be obtained is akasakaakasaka, whose length is 14.

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

// include
// ------------------------------------------------
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

// func
// ------------------------------------------------
int CalcSumOfDigit(int n);      // 各桁の和を計算する。
string upper(string str);       // 英字を大文字に変換する。
string lower(string str);       // 英字を小文字に変換する。

// define
// ------------------------------------------------
#define all(a)        (a).begin(),(a).end()
#define rall(a)       (a).rbegin(),(a).rend()
#define sz(a)         int((a).size())
#define rep(i,n)      for(int(i)=0;(i)<(n);(i)++)
#define repe(i,n)     for(int(i)=0;(i)<=(n);(i)++)
#define vsort(v)      sort((v).begin(),(v).end())
#define rvsort(v)     sort(rall((v)))
#define vi            vector<int>
#define GCD(a,b)      __gcd((a),(b))
#define LCM(a,b)      (a)/GCD((a),(b))*(b)
#define kiriage(a,b)  ((a)+(b)-1)/(b)

const int INF = 1e9;

// code
// ------------------------------------------------
int main() {

  string s,a,b;
  cin >> s;

  for(int i = 1; i <= sz(s) - 2;i++)
  {
    s = s.substr(0,sz(s) - 1);
    a = s.substr(0,sz(s) / 2 );
    b = s.substr(sz(s) / 2 );
    if(a == b)
    {
      cout << sz(s) << endl;
      return 0;
    }
  }

  return 0;
}
// funcの実体
// ------------------------------------------------
int CalcSumOfDigit(int n)
{
  int s = 0;
  while(n)
  {
    s += n % 10;
    n = n / 10;
  }
  return s;
}

string upper(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(97 <= *itr && *itr <= 122)
    {
      *itr = *itr - 32;
    }
  }
  return str;
}

string lower(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(65 <= *itr && *itr <= 90)
    {
      *itr = *itr + 32;
    }
  }
  return str;
}