#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2")
using namespace std;
auto clk = clock();
const long double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const long double eps = 1e-10;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const int MOD = 998244353;
const long long MOD2 = (long long)MOD * (long long)MOD;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
const int mod = 1000000007;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int add(int x, int y, int z) { return add(add(x, y), z); }
inline int sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int mul(int x, int y) { return 1LL * x * y % mod; }
inline int mul(int x, int y, int z) { return mul(mul(x, y), z); }
const string yes = "YES";
const string no = "NO";
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string str;
  cin >> str;
  int n = str.length();
  vector<char> ans(n);
  map<char, int> um;
  sort(str.begin(), str.end());
  for (int i = 0; str[i]; i++) {
    um[str[i]]++;
  }
  int oddc = 0;
  vector<char> vc;
  for (auto &x : um) {
    if (x.second % 2 == 1) {
      vc.push_back(x.first);
      x.second--;
    }
  }
  oddc = vc.size();
  if (oddc != 1)
    for (int i = 0; i < oddc / 2; i++) {
      vc[oddc - i] = vc[i];
      um[vc[i]] += 2;
    }
  deque<char> dq;
  if (oddc % 2 == 1) {
    dq.push_back(vc[oddc / 2]);
  }
  for (int i = 'z'; i >= 'a'; i--) {
    while (um[i] != 0) {
      dq.push_back(i);
      dq.push_front(i);
      um[i] -= 2;
    }
  }
  while (!dq.empty()) {
    cout << dq.front();
    dq.pop_front();
  }
  cout << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 GCC optimize("Ofast")
#pragma GCC target("avx,avx2")
using namespace std;
auto clk = clock();
const long double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const long double eps = 1e-10;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const int MOD = 998244353;
const long long MOD2 = (long long)MOD * (long long)MOD;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
const int mod = 1000000007;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int add(int x, int y, int z) { return add(add(x, y), z); }
inline int sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int mul(int x, int y) { return 1LL * x * y % mod; }
inline int mul(int x, int y, int z) { return mul(mul(x, y), z); }
const string yes = "YES";
const string no = "NO";
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string str;
  cin >> str;
  int n = str.length();
  vector<char> ans(n);
  map<char, int> um;
  sort(str.begin(), str.end());
  for (int i = 0; str[i]; i++) {
    um[str[i]]++;
  }
  int oddc = 0;
  vector<char> vc;
  for (auto &x : um) {
    if (x.second % 2 == 1) {
      vc.push_back(x.first);
      x.second--;
    }
  }
  oddc = vc.size();
  if (oddc != 1)
    for (int i = 0; i < oddc / 2; i++) {
      vc[oddc - i] = vc[i];
      um[vc[i]] += 2;
    }
  deque<char> dq;
  if (oddc % 2 == 1) {
    dq.push_back(vc[oddc / 2]);
  }
  for (int i = 'z'; i >= 'a'; i--) {
    while (um[i] != 0) {
      dq.push_back(i);
      dq.push_front(i);
      um[i] -= 2;
    }
  }
  while (!dq.empty()) {
    cout << dq.front();
    dq.pop_front();
  }
  cout << '\n';
  return 0;
}

```