You are given a non-empty string s consisting of lowercase English letters. You have to pick exactly one non-empty substring of s and shift all its letters 'z' <image> 'y' <image> 'x' <image> 'b' <image> 'a' <image> 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.

What is the lexicographically minimum string that can be obtained from s by performing this shift exactly once?

Input

The only line of the input contains the string s (1 ≤ |s| ≤ 100 000) consisting of lowercase English letters.

Output

Print the lexicographically minimum string that can be obtained from s by shifting letters of exactly one non-empty substring.

Examples

Input

codeforces


Output

bncdenqbdr


Input

abacaba


Output

aaacaba

Note

String s is lexicographically smaller than some other string t of the same length if there exists some 1 ≤ i ≤ |s|, such that s1 = t1, s2 = t2, ..., si - 1 = ti - 1, and si < ti.

import java.util.Scanner;

public class Problem {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String input = sc.next();

		if (!input.contains("a")) {
			char[] inputChars = input.toCharArray();
			for (int i = 0; i < inputChars.length; i++) {
				inputChars[i] = (char) (inputChars[i] - 1);
			}
			System.out.println(String.valueOf(inputChars));
		} else {
			String[] subStrs = input.split("a");
			if (subStrs.length == 0) {
				char[] shitChars = input.toCharArray();
				shitChars[input.length() - 1] = 'z';
				System.out.println(String.valueOf(shitChars));
				return;
			}
			int index = 0;
			for (int i = 0; i < subStrs.length; i++) {
				if (subStrs[i].length() != 0) {
					index = i;
					break;
				}
			}
			char[] inputChars = input.toCharArray();
			char[] shitChars = subStrs[index].toCharArray();

			int start = input.indexOf(String.valueOf(subStrs[index]));

			for (int i = start; i < start + shitChars.length; i++) {
				inputChars[i] = (char) (inputChars[i] - 1);
			}

			System.out.println(String.valueOf(inputChars));

		}

	}

}
