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.*;
import java.util.*;

public class CTask {

    public static void main(String[] args) throws IOException {
        MyInputReader in = new MyInputReader(System.in);
        char[] str = in.next().toCharArray();
        int pos = 0;
        while (pos < str.length) {
            if (str[pos] != 'a') {
                break;
            }
            pos += 1;
        }
        if (pos == str.length) {
            str[str.length - 1] = 'z';
        }
        while (pos < str.length && str[pos] != 'a') {
            str[pos] -= 1;
            pos += 1;
        }
        System.out.println(new String(str));
    }



    static class MyInputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public MyInputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }

}