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;
const long long OO = 1e8;
const double EPS = (1e-9);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void PLAY() {
  cout << fixed << setprecision(4);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int up(vector<int> &v, int num) {
  int lo = 0, hi = ((int)((v).size())) - 1;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (v[mid] <= num)
      lo = mid + 1;
    else if (v[mid] > num)
      hi = mid - 1;
  }
  if (lo >= ((int)((v).size()))) return -1;
  return v[lo];
}
long long down(vector<long long> &v, long long num) {
  long long lo = 0, hi = ((int)((v).size())) - 1;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if (v[mid] < num)
      lo = mid + 1;
    else if (v[mid] >= num)
      hi = mid - 1;
  }
  if (hi < 0) return -1;
  return v[hi];
}
bool all_chars_same(string testStr) {
  return testStr.find_first_not_of(testStr[0]) == string::npos;
}
void removeCharsFromString(string &str, char *charsToRemove) {
  for (unsigned int i = 0; i < strlen(charsToRemove); ++i) {
    str.erase(remove(str.begin(), str.end(), charsToRemove[i]), str.end());
  }
}
bool compare(const pair<int, int> &i, const pair<int, int> &j) {
  return i.first < j.first;
}
int f(string s) {
  long long int n = s.size(), num = 0;
  static long long int cnt = 0;
  string str;
  if (n == 1) return cnt;
  cnt++;
  for (int i = 0; i < (int)(n); ++i) {
    num += s[i] - '0';
  }
  stringstream o;
  o << num;
  str = o.str();
  return f(str);
}
template <typename Container>
typename Container::const_iterator constify(const Container &,
                                            typename Container::iterator it) {
  return it;
}
template <typename T>
std::pair<bool, int> findInVector(
    const std::vector<unsigned int> &vecOfElements, const T &element) {
  std::pair<bool, unsigned int> result;
  vector<unsigned int>::const_iterator it =
      find(vecOfElements.begin(), vecOfElements.end(), element);
  if (it != vecOfElements.end()) {
    result.second = distance(vecOfElements.begin(), it);
    result.first = true;
  } else {
    result.first = false;
    result.second = -1;
  }
  return result;
}
long long memo[101][101];
long long C(int a, int b) {
  if (b == 0) return 1;
  if (a == b) return 1;
  if (memo[a][b] != -1) return memo[a][b];
  memo[a][b] = C(a - 1, b) + C(a - 1, b - 1);
  return memo[a][b];
}
void files() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
int main() {
  string s1, s2;
  cin >> s1;
  s2 = s1;
  reverse(((s2).begin()), ((s2).end()));
  s1 += s2;
  cout << s1;
  return 0;
}