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;
int L = 2e5 + 5;
const int MAXN = 2e5 + 5;
int M = 1073741824;
struct cmp {
  bool operator()(const std::pair<int, int> &left,
                  const std::pair<int, int> &right) {
    return ((abs(left.first) <= abs(right.first)));
  }
};
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
unsigned long long gcd1(unsigned long long a, unsigned long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
bool is_prime(int a) {
  int id = 0;
  for (int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) return false;
  }
  return true;
}
long long check(long long a) {
  long long cnt = 0;
  for (int i = 1; i <= sqrt(a); i++) {
    if (a % i == 0) {
      cnt++;
    }
  }
  long long g = sqrt(a);
  if (g * g == a) {
    cnt = (2 * cnt - 1);
  } else
    cnt = 2 * cnt;
  return cnt;
}
void recur(string &s, int id) {
  if (id == s.size() - 1)
    return;
  else {
    s.insert(id + 1, "*");
    id += 1;
  }
  recur(s, id + 1);
}
long long gcd2(long long a, long long b) { return b ? gcd2(b, a % b) : a; }
unsigned long long int GCD(unsigned long long int a, unsigned long long int b) {
  unsigned long long int c = a % b;
  if (c == 0) return b;
  return GCD(b, c);
}
int accept() {
  string s;
  cin >> s;
  string r = s;
  reverse(s.begin(), s.end());
  cout << r << s << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    accept();
  }
  return 0;
}