Problem: Shil is your new boss and he likes palindromes very much.  Palindrome is a string that can be read the same way in either direction, from the left to the right and from the right to the left. (ex. madam , aabaa, racecar)  

Given a string S , beautiful Palindrome is a lexicographical minimum palindrome that can be formed by rearranging all the characters of string S. In order to please your boss, find a beautiful Palindrome that can be formed with help of string S.  

String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages​​.

Input:
Only line of input contains string S. All the letters of this string will be in lower letters('a' - 'z').

Output:
Output lexicographical minimum Palindrome that can be formed by rearranging all the letters of string S. If no such Palindrome exist for given  input, print -1.

Constraints: 
1≤|S|≤100000

SAMPLE INPUT
aabcc

SAMPLE OUTPUT
acbca

Explanation

After rearranging all the characters , all the palindromes that can be formed are cabac and acbca. Out of these two lexicographical minimum one is acbca.

Solution (in PYTHON):

Solution:
b=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
s=raw_input()
a=[0]*30
le=len(s)
for i in range(0,le):
	a[ord(s[i])-ord('a')]=a[ord(s[i])-ord('a')]+1
final=""
temp=""
for i in range(0,26):
	if(a[i]%2==0):
		final=final+b[i]*(a[i]/2)
	else:
		final=final+b[i]*(a[i]/2)
		temp=temp+b[i]
		
final=final+temp+final[::-1]	
flag=0
for i in range(0,le):
	if(final[i]!=final[le-i-1]):
		flag=1
		break
if(flag):
	print -1
else:
	print final