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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;

struct Manacher {
  template <typename T>
  Manacher(const T &s) {
    T str;
    int n = s.size();
    REP(i, n) {
      str.push_back(s[i]);
      if (i + 1 < n) str.push_back('$');
    }
    n = str.size();
    radius.resize(n);
    int j = 1;
    for (int i = 0; i < n;) {
      while (i - j >= 0 && i + j < n && str[i - j] == str[i + j]) ++j;
      radius[i] = j;
      int k = 1;
      while (i - k >= 0 && i + k < n && k + radius[i - k] < j) {
        radius[i + k] = radius[i - k];
        ++k;
      }
      i += k;
      j -= k;
    }
  }

  int odd(int idx) { return (radius[idx * 2] + 1) / 2; }

  int even(int idx) { return radius[idx * 2 + 1] / 2; }

  bool is_palindrome(int left, int right) {
    int mid = (left + right - 1) / 2;
    return (((right - left) & 1) ? odd(mid) * 2 - 1 : even(mid) * 2) >= right - left;
  }

private:
  vector<int> radius;
};

int main() {
  string s; cin >> s;
  int n = s.length();
  Manacher m(s);
  cout << (m.is_palindrome(0, n) && m.is_palindrome(0, (n - 1) / 2) && m.is_palindrome((n + 1) / 2, n) ? "Yes\n" : "No\n");
  return 0;
}