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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    dead_code
)]
//use permutohedron::LexicalPermutation as _;
//use proconio::input;
//use proconio::marker::Chars;
//use proconio::marker::Usize1;
//use std::collections::HashSet;
//use std::collections::HashMap;
//use std::collections::BTreeMap;
//use std::cmp::*;
//use std::num;

fn main() {
    proconio::input! {
        S:proconio::marker::Chars
    }
    let l = S.len();
    for i in 0..(l - 1) / 2 {
        if S[i] != S[l - 1 - i] {
            println!("No");
            return;
        }
    }
    for i in 0..(l - 1) / 4 + 1 {
        if S[i] != S[(l - 1) / 2 - 1 - i] {
            println!("No");
            return;
        }
    }
    println!("Yes");
}