Problem:
Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <cstring>
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <cmath>
#include <numeric>
#include <float.h>
using namespace std;
// repマクロの定義
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
// allマクロの定義
#define all(v) v.begin(), v.end()
typedef unsigned long long ll;

// 最大公約数求める関数
int my_gcd(int n , int m){
  if(n < m){
    int tmp = n;
    n = m;
    m = tmp;
  }
  int l  = n % m;
  while(l != 0){
    n = m;
    m = l;
    l = n % m;
}
return m;
}

//char⇒int関数
 int ctoi(char c) {
 	switch (c) {
		case '0': return 0;
		case '1': return 1;
		case '2': return 2;
		case '3': return 3;
		case '4': return 4;
		case '5': return 5;
		case '6': return 6;
		case '7': return 7;
		case '8': return 8;
		case '9': return 9;
		default: return 0;
	}
}



int main(){
  string S;
  cin >> S;
  int N = S.size();
  vector<char> A(50),B(50);
  rep(i,N/2){
    A[i] = S[i];
    B[i+(N/2)+1] = S[i+(N/2)+1];
  }
  bool ans = true;
  rep(i,N/2){
    if(A[i] == S[N/2 - i - 1] && B[i+ 1 + (N/2)] == S[N -i - 1]){
    }
    else ans = false;
  }
  if(ans) cout << "Yes" << endl;
  else cout << "No" << endl;
  }