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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FS in = new FS();
		PrintWriter out = new PrintWriter(System.out);
		String s = in.next();
		StringBuilder sb = new StringBuilder();
		boolean started = false;
		boolean ended = false;
		for(char c : s.toCharArray()){
			if(ended){
				sb.append(c);
				continue;
			}
			if(c=='a'){
				if(started){
					ended = true;
				}
				sb.append(c);
				continue;
			}
			started = true;
			sb.append((char)((int)c-1));
		}
		if(!started){
			sb.replace(sb.length()-1, sb.length(), "z");
		}
		out.print(sb.toString());
		out.close();
	}
	private static class FS {
		BufferedReader br;
		StringTokenizer st;
		
		public FS () {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch (Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
}
}
