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.StringTokenizer;

/**
 * 708A
 * θ(n) time
 * θ(n) space
 *
 * @author artyom
 */
public class _708A implements Runnable {
    private BufferedReader in;
    private StringTokenizer tok;

    private Object solve() throws IOException {
        char[] s = nextToken().toCharArray();
        int n = s.length, i = 0;
        while (i < n && s[i] == 'a') {
            i++;
        }
        if (i == n) {
            s[n - 1] = 'z';
        } else {
            while (i < n && s[i] != 'a') {
                s[i] = (char) (s[i] - 1);
                i++;
            }
        }
        return new String(s);
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new _708A().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tok = null;
            System.out.print(solve());
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

    private String nextToken() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
}