#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const int INF = (int)(1e9);
const long long INF64 = (long long)(INF) * (long long)(INF);
const double eps = 1e-9;
const double pi = double(3.1415926535897932384626433832795);
inline bool in(int i, int j, int n, int m) {
  return i >= 1 && i <= n && j >= 1 && j <= m;
}
inline int myrand() { return (rand() ^ (rand() << 15)); }
const int dx[] = {0, -1, 1, 0};
const int dy[] = {-1, 0, 0, 1};
template <typename T1, typename T2>
inline ostream &operator<<(ostream &out, const pair<T1, T2> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <typename T>
inline void outArr(const T a[], int n) {}
template <typename T>
inline void out(const T &a) {}
template <typename T1, typename T2>
inline void out(const pair<T1, T2> &a) {}
template <typename T>
inline void out(const vector<T> &a) {}
template <typename T>
inline void out(const set<T> &a) {}
const int N = 2e5 + 555;
string s;
inline void gen() { return; }
inline bool read() {
  cin >> s;
  return true;
}
inline bool check(const string &s) {
  int l = 0, r = (int)((s).size()) - 1;
  while (l < r) {
    if (s[l] != s[r]) return false;
    l++, r--;
  }
  return true;
}
int cnt[256];
inline void solve() {
  int n = (int)((s).size());
  for (int i = 0; i < (int)(n); i++) cnt[s[i]]++;
  int odd = 0;
  vector<int> Odd;
  for (int i = 0; i < (int)(256); i++)
    if (cnt[i] & 1) Odd.push_back(i);
  if ((int)((Odd).size()) > 1) {
    int l = 0, r = (int)((Odd).size()) - 1;
    while (l < r) {
      cnt[Odd[l]]++;
      cnt[Odd[r]]--;
      l++, r--;
    }
  }
  string ans;
  ans.erase();
  int mid = -1;
  for (int i = 0; i < (int)(256); i++)
    if (cnt[i] & 1) mid = i;
  for (int i = 0; i < (int)(256); i++) {
    for (int j = 0; j < (int)(cnt[i] / 2); j++) ans += char(i);
  }
  if (mid > 0) ans += char(mid);
  for (int i = 255; i >= 0; i--) {
    for (int j = 0; j < (int)(cnt[i] / 2); j++) ans += char(i);
  }
  assert(check(ans));
  cout << ans << endl;
  return;
}
int main() {
  cout << setprecision(10) << fixed;
  cerr << setprecision(10) << fixed;
  srand(int(time(NULL)));
  int T = 1;
  for (int i = 0; i < (int)(T); i++) {
    assert(read());
    solve();
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo", "r" and "mikhailrubinchikkihcniburliahkim" are palindroms, but strings "abb" and "ij" are not.

You are given string s consisting of lowercase Latin letters. At once you can choose any position in the string and change letter in that position to any other lowercase letter. So after each changing the length of the string doesn't change. At first you can change some letters in s. Then you can permute the order of letters as you want. Permutation doesn't count as changes. 

You should obtain palindrome with the minimal number of changes. If there are several ways to do that you should get the lexicographically (alphabetically) smallest palindrome. So firstly you should minimize the number of changes and then minimize the palindrome lexicographically.

Input

The only line contains string s (1 ≤ |s| ≤ 2·105) consisting of only lowercase Latin letters.

Output

Print the lexicographically smallest palindrome that can be obtained with the minimal number of changes.

Examples

Input

aabc


Output

abba


Input

aabcd


Output

abcba

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const int INF = (int)(1e9);
const long long INF64 = (long long)(INF) * (long long)(INF);
const double eps = 1e-9;
const double pi = double(3.1415926535897932384626433832795);
inline bool in(int i, int j, int n, int m) {
  return i >= 1 && i <= n && j >= 1 && j <= m;
}
inline int myrand() { return (rand() ^ (rand() << 15)); }
const int dx[] = {0, -1, 1, 0};
const int dy[] = {-1, 0, 0, 1};
template <typename T1, typename T2>
inline ostream &operator<<(ostream &out, const pair<T1, T2> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <typename T>
inline void outArr(const T a[], int n) {}
template <typename T>
inline void out(const T &a) {}
template <typename T1, typename T2>
inline void out(const pair<T1, T2> &a) {}
template <typename T>
inline void out(const vector<T> &a) {}
template <typename T>
inline void out(const set<T> &a) {}
const int N = 2e5 + 555;
string s;
inline void gen() { return; }
inline bool read() {
  cin >> s;
  return true;
}
inline bool check(const string &s) {
  int l = 0, r = (int)((s).size()) - 1;
  while (l < r) {
    if (s[l] != s[r]) return false;
    l++, r--;
  }
  return true;
}
int cnt[256];
inline void solve() {
  int n = (int)((s).size());
  for (int i = 0; i < (int)(n); i++) cnt[s[i]]++;
  int odd = 0;
  vector<int> Odd;
  for (int i = 0; i < (int)(256); i++)
    if (cnt[i] & 1) Odd.push_back(i);
  if ((int)((Odd).size()) > 1) {
    int l = 0, r = (int)((Odd).size()) - 1;
    while (l < r) {
      cnt[Odd[l]]++;
      cnt[Odd[r]]--;
      l++, r--;
    }
  }
  string ans;
  ans.erase();
  int mid = -1;
  for (int i = 0; i < (int)(256); i++)
    if (cnt[i] & 1) mid = i;
  for (int i = 0; i < (int)(256); i++) {
    for (int j = 0; j < (int)(cnt[i] / 2); j++) ans += char(i);
  }
  if (mid > 0) ans += char(mid);
  for (int i = 255; i >= 0; i--) {
    for (int j = 0; j < (int)(cnt[i] / 2); j++) ans += char(i);
  }
  assert(check(ans));
  cout << ans << endl;
  return;
}
int main() {
  cout << setprecision(10) << fixed;
  cerr << setprecision(10) << fixed;
  srand(int(time(NULL)));
  int T = 1;
  for (int i = 0; i < (int)(T); i++) {
    assert(read());
    solve();
  }
  return 0;
}

```