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.*;
import java.io.*;

public class Main {
	static class Reader {
		BufferedReader br; StringTokenizer st;

		public Reader() { br = new BufferedReader(new InputStreamReader(System.in)); }

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try { st = new StringTokenizer(br.readLine());
				} catch (IOException e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}

		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}

	public static void main(String[] args) {
		Reader r = new Reader();

		String s = r.next();
		String o = "";

		int done = 0;
		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			if (c != 'a' && done < 1) {
				while (i < s.length()) {
					done = 1;
					c = s.charAt(i);
					i++;
					if (c == 'a') {
						o += c;
						done = 2;
						break;
					}
					o += (char)(c - 1);
				}
				i--;
			} else {
				o += c;
			}
		}

		if (done == 0) o = s.substring(0, s.length() - 1) + 'z';
		System.out.println(o);
	
	}

}
