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
#include "bits/stdc++.h"
using namespace std;

int main() {
	string s;
	cin >> s;

	string ans = "-1";
	if (s.size() < 26) {

		vector<bool> vec(26, false);
		for (auto& ch : s) {
			vec[ch - 'a'] = true;
		}

		ans = s;
		for (int i = 0; i < 26; i++) {
			if (vec[i]) continue;
			ans += ('a' + i);
			break;
		}
	}
	else {
		for (int i = 25; i > 0; i--) {
			if (s[i] > s[i - 1]) {
				char ch = '~';
				for (int j = i - 1; j < 26; j++) {
					if (s[i - 1] < s[j]) ch = min(ch, s[j]);
				}
				ans = s.substr(0, i - 1) + ch;
				break;
			}
		}
	}

	cout << ans << endl;
}