Problem: Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.

Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.

Now Pari asks you to write a program that gets a huge integer n from the input and tells what is the n-th even-length positive palindrome number?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10100 000).

Output

Print the n-th even-length palindrome number.

Examples

Input

1


Output

11


Input

10


Output

1001

Note

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long base, long long exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base);
    base = (base * base);
    exp /= 2;
  }
  return res;
}
long long mod(long long a, long long b) { return (a % b + b) % b; }
long long powerm(long long base, long long exp, long long mod) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = (ans * base) % mod;
    exp >>= 1, base = (base * base) % mod;
  }
  return ans;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  t = 1;
  while (t--) {
    string s;
    cin >> s;
    cout << s;
    for (long long i = s.size() - 1; i >= 0; i--) cout << s[i];
    cout << '\n';
  }
  return 0;
}