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.*;
import java.text.*;
import java.math.*;

public class main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static PrintWriter pw = new PrintWriter(System.out);
    public static String line;
    public static StringTokenizer st;
    public static ArrayList<ArrayList<Integer>> adjList;
    public static int[] dr = {-1, 0, 1, 0, -1, 1, 1, -1};
    public static int[] dc = {0, 1, 0, -1, 1, 1, -1, -1};

    public static void main(String[] args) throws IOException{
        char[] A = br.readLine().toCharArray();

        int i = 0;
        boolean f = false;
        while(i < A.length && A[i] == 'a'){
            i++;
        }

        for(; i < A.length; i++){
            if(A[i] != 'a'){
                A[i] = (char) ((int)A[i] - 1);
                f = true;
            } else{
                break;
            }
        }

        if(!f){
            A[A.length-1] = 'z';
        }

        pw.print(new String(A) +"\n");

        pw.close();
        br.close();
    }
}
