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 A implements Runnable {

    private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tok = new StringTokenizer("");

    private void init() throws FileNotFoundException {
        Locale.setDefault(Locale.US);
        String fileName = "";
        if (ONLINE_JUDGE && fileName.isEmpty()) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            try {
                if (fileName.isEmpty()) {
                    in = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter("output.txt");
                } else {
                    in = new BufferedReader(new FileReader(fileName + ".in"));
                    out = new PrintWriter(fileName + ".out");
                }
            } catch (Throwable t) {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            }
        }
    }

    String readString() {
        while (!tok.hasMoreTokens()) {
            try {
                tok = new StringTokenizer(in.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return tok.nextToken();
    }

    int readInt() {
        return Integer.parseInt(readString());
    }

    long readLong() {
        return Long.parseLong(readString());
    }

    double readDouble() {
        return Double.parseDouble(readString());
    }

    int[] readIntArray(int size) {
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = readInt();
        }
        return a;
    }

    public static void main(String[] args) {
        //new Thread(null, new _Solution(), "", 128 * (1L << 20)).start();
        new A().run();
    }

    long timeBegin, timeEnd;

    void time() {
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }

    @Override
    public void run() {
        try {
            timeBegin = System.currentTimeMillis();
            init();
            solve();
            out.close();
            time();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    private void solve() throws IOException {
        char[] s = readString().toCharArray();
        for (int i = 0; i < s.length; i++) {
            if (s[i] != 'a') {
                for (int j = i; j < s.length; j++) {
                    if (s[j] == 'a') {
                        break;
                    }
                    s[j]--;
                }
                for (int k = 0; k < s.length; k++) {
                    out.print(s[k]);
                }
                return;
            }
        }
        for (int i = s.length - 1; i >= 0; i--) {
            if (s[i] == 'a') {
                s[i] = 'z';
                for (int k = 0; k < s.length; k++) {
                    out.print(s[k]);
                }
                return;
            }
        }
    }
}
