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.util.*;
import java.io.*;

public class C {
    char prev(char c) {
        if (c != 'a')
            return (char)((int)c - 1);
        return 'z';
    }

    void solve() throws IOException {
        String s = next();
        int indStart, indEnd;
        indStart = 0;
        while (indStart < s.length() - 1 && s.charAt(indStart) == 'a')
            indStart++;

        indEnd = indStart + 1;
        while (indEnd < s.length() && s.charAt(indEnd) != 'a')
            indEnd++;


        StringBuilder sb = new StringBuilder(s);
        for (int i = indStart; i < indEnd; i++) {
            sb.setCharAt(i, prev(sb.charAt(i)));
        }

        output.println(sb.toString());
    }

    public static void main(String[] args) throws IOException {
        new C();
    }

    StringTokenizer st = null;
    BufferedReader br;
    PrintWriter output;

    C() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        output = new PrintWriter(System.out);
        solve();
        br.close();
        output.close();
    }

    String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());

        return st.nextToken();
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

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

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
}