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 static java.lang.Math.*;

public class Main {
    public static class Solver {
        Scanner in = new Scanner(System.in);

        <T> void println(T x) {
            System.out.println(x);
        }

        <T> void print(T x) {
            System.out.print(x);
        }

        int d(int p, int q) {
            return abs(p - q);
        }

        char shift(char c) {
            return c == 'a' ? 'z' : (char)((int)c - 1);
        }

        void solve() {
            String s = in.nextLine();
            int i = 0;
            int n = s.length();
            while (i < n && s.charAt(i) == 'a') {
                print(i == n - 1 ? shift(s.charAt(i)) : s.charAt(i));
                i++;
            }

            while (i < n && s.charAt(i) != 'a') {
                print(shift(s.charAt(i)));
                i++;
            }

            while (i < n) {
                print(s.charAt(i));
                i++;
            }
        }
    }

    public static void main(String[] args) {
        new Solver().solve();
    }
}