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: Accepted)

#include "./bits/stdc++.h"

using namespace std;

bool check(string s) {
    int n = s.length();
    int i = 0;
    while (1) {
        if (s[i] != s[n - 1 - i]) return false;
        if (n - (i + 1) * 2 <= 1) return true;
        i++; 
    }
}

int main() {
    string S;
    cin >> S;

    // 回文の切れ目
    int N = S.length();
    int b1 = (N - 1) / 2;
    int b2 = (N + 3) / 2;

    string s1, s2, s3;

    for (int i = 0; i < b1; i++) s1 += S[i];
    for (int i = b1; i < b2 - 1; i++) s2 += S[i];
    for (int i = b2 - 1; i < N; i++) s3 += S[i];

    bool result = check(s1) && check(s2) && check(s3) && s1 == s3;

    cout << (result ? "Yes" : "No") << endl;
    return 0;
}