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.*;
import java.util.stream.IntStream;

public class A {
    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        char[] s = in.next().toCharArray();
        int i = 0;
        while (i < s.length && s[i] == 'a') {
            i++;
        }
        if (i == s.length) {
            s[s.length - 1] = 'z';
            out.println(s);
        } else {
            while (i < s.length && s[i] != 'a') {
                s[i]--;
                i++;
            }
            out.println(s);
        }
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);
            Locale.setDefault(Locale.US);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

    public static void main(String[] arg) {
        new A().run();
    }
}