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)

/*
If you want to shine like a sun, first burn like a sun.
*/

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds;
using namespace std;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define ll long long int
#define ld long double
#define vi vector<int>
#define vl vector<ll>
#define v2i vector<vector<int>>
#define v2l vector<vector<ll>>
#define ppi pair<int,int>
#define ppl pair<ll,ll>
#define vpi vector<ppi>
#define vpl vector<ppl>
#define all(x) x.begin(),x.end()
#define ii int,int 
#define ff first    
#define ss second
#define forn(i,a,b) for(int i=a;i<b;i++)
#define forr(i,a,b) for(int i=a;i>=b;i--)
#define forv(i,m) for(auto i : m)
#define imx INT_MAX
#define imn INT_MIN
#define inf 9000000000000000000
#define minf -inf
#define endl "\n"
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define sze size()
#define rvs reverse 
#define itr iterator 
#define pre cout<<fixed<<setprecision(10);
#define umap unordered_map
#define uset unordered_set
#define pi 3.141592653589793
#define MAXN 100005
#define test1(x) cout << #x " = " << x << endl;
#define test2(x,y) cout << #x " = " << x << " " << #y " = " << y << endl;
const ll infinity = 9e18;

/*
#define oset tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define MODM(a,b,md) ((a%md)*(b%md))%md
#define MODA(a,b,md) ((a%md)+(b%md))%md
#define ull unsigned long long int
*/

/*
bool compare(int a,int b) {return a > b;}
bool compare1(ppi a,ppi b) {return a.ff > b.ff;}
bool compare2(ppi a,ppi b) {return a.ss < b.ss;}
ll Npower(ll a,ll b) {ll product = 1; while(b){if(b&1){product = product*a;}a = a*a;b = b>>1;} return product;}
ll power(ll a,ll b,ll md) {ll product = 1; while(b){if(b&1){product = MODM(product,a,md);}a = MODM(a,a,md);b = b>>1;} return product%md;}
bool isprime(ll n){if(n < 2) return 0; ll i = 2; while(i*i <= n){if(n%i == 0) return 0; i++;} return 1;}
ll GCD(ll a,ll b) {return b==0 ? a:GCD(b,a%b);}
ll LCM(ll a,ll b) {return (a/GCD(a,b))*b; }
*/

bool isPalindrome(string s)
{

    forn(i,0,s.sze/2)
    {
        if(s[i] != s[s.sze-i-1]) return false;
    }

    return true;
}


int main()
{
    fast

/*
    #ifndef ONLINE_JUDGE
        freopen("input.txt",  "r",  stdin);
        freopen("output.txt", "w", stdout);
    #endif
//*/

    string s;
    cin>>s;

    int n = s.sze;
    if(isPalindrome(s) && isPalindrome(s.substr(0,(n-1)/2)) && isPalindrome(s.substr((n+1)/2))) cout<<"Yes";
    else cout<<"No";







    return 0;
}