Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

* 1 \leq |S| \leq 26
* S is a diverse word.

Input

Input is given from Standard Input in the following format:


S


Output

Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.

Examples

Input

atcoder


Output

atcoderb


Input

abc


Output

abcd


Input

zyxwvutsrqponmlkjihgfedcba


Output

-1


Input

abcdefghijklmnopqrstuvwzyx


Output

abcdefghijklmnopqrstuvx
s=list(raw_input())
l=len(s)-1
if len(s)<26:
    al = {}
    for i in s:
        al[i] = 1
    while (l >= 0):
        for i in range(97, 97 + 26):
            x = chr(i)
            if al.get(x, -1) == -1 :
                ans = s[:l + 1] + [x]
                if len(s) == 26:
                    ans[-2] = ''
                print ''.join(ans)
                exit()
        al[s[l]] = -1
        l -= 1
    print -1
else:
    al={}
    for i in s:
        al[i]=1
    while(l>=0):
        for i in range(97,97+26):
            x=chr(i)
            if al.get(x,-1)==-1 and x>s[l]:
                ans=s[:l+1] +[x]
                if len(s)==26:
                    ans[-2]=''
                print ''.join(ans)
                exit()
        al[s[l]]=-1
        l-=1
    print -1





