CodeForces - 600C Make Palindrome 贪心

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 ins. 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.

Example

Input

aabc

Output

abba

Input

aabcd

Output

abcba

题目大意：输入一串字符串，改变其中的字母或调整顺序，最后使得输出的字符串为字典序最小的回文，且要求变换次数最小（调准顺序不算）

思路：先使用一个数组，用来记录每个字符出现的次数，再用一个数组来记录个数为奇数的字母，然后从头尾两端向中间遍历，大的减一小的加一。最后再看为奇数字母的个数，若为偶数则按字母从小到大排序即可，若为奇数，则把最后为奇数个的字母放在中间，其余的按从小到大排序。

#include
#include
char str[200000];
int main()
{
	int n = 0;
	int re[26] = { 0 }, ch[26] = {0};
	
	int len;
	scanf("%s",str);
	len = strlen(str);
	for (int i = 0; i < len; i++)
	{
		re[str[i]-'a']++;
	}
	for (int i = 0; i < 26; i++)
	{
		if (re[i] % 2 != 0)
		{
			ch[n++] = i;
		}
	}
	for (int i = 0, j = n-1; i < j; i++, j--)
	{
		re[ch[i]]++;
		re[ch[j]]--;
	}
	if (n % 2 != 0)
	{
		str[len / 2] = ch[n / 2] + 'a';
	}
	for (int i = 0, j = len-1,h=0; i <= j; i++, j--)
	{
		for (; h < 26; h++)
		{
			if (re[h] >= 2)
			{
				re[h] -= 2;
				str[i] = h+'a';
				str[j] = h+'a';
				break;
			}
		}
	}
	printf("%s", str);

	return 0;
}