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.math.BigInteger;
import java.util.*;

public class Solution {
    static MyScanner sc;
    private static PrintWriter out;
    static long M2 = 1_000_000_000L + 7;

    public static void main(String[] s) throws Exception {
        StringBuilder stringBuilder = new StringBuilder();
//        stringBuilder.append("100000 1000000000 10001 100000 ");
//        for (int i = 1; i < 100000; i++) {
//            stringBuilder.append(" 10000 100000 ");
//        }
        if (stringBuilder.length() == 0) {
            sc = new MyScanner(System.in);
        } else {
            sc = new MyScanner(new BufferedReader(new StringReader(stringBuilder.toString())));
        }

        out = new PrintWriter(new OutputStreamWriter(System.out));
        initData();
        solve();
        out.flush();
    }


    private static void initData() {

    }

    private static void solve() throws IOException {
        String s = sc.next();
        char[] pp = s.toCharArray();
        boolean v = false;
        for (int i = 0; i < pp.length; i++) {
            if (pp[i] == 'a') {
                if (v) break;
            } else {
                v = true;
                pp[i]--;
            }
        }
        if (!v) pp[pp.length - 1] = 'z';
        out.println(new String(pp));
    }


    private static void solveT() throws IOException {
        int t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    private static long gcd(long l, long l1) {
        if (l > l1) return gcd(l1, l);
        if (l == 0) return l1;
        return gcd(l1 % l, l);
    }

    private static long pow(long a, long b, long m) {
        if (b == 0) return 1;
        if (b == 1) return a;
        long pp = pow(a, b / 2, m);
        pp *= pp;
        pp %= m;
        return (pp * (b % 2 == 0 ? 1 : a)) % m;
    }


    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        MyScanner(BufferedReader br) {
            this.br = br;
        }

        public MyScanner(InputStream in) {
            this(new BufferedReader(new InputStreamReader(in)));
        }

        void findToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
        }

        String next() {
            findToken();
            return st.nextToken();
        }

        Integer[] nab(int n) {
            Integer[] k = new Integer[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        int[] na(int n) {
            int[] k = new int[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        long[] nl(int n) {
            long[] k = new long[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.nextLong();
            }
            return k;
        }

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

        int fi() {
            String t = next();
            int cur = 0;
            boolean n = t.charAt(0) == '-';
            for (int a = n ? 1 : 0; a < t.length(); a++) {
                cur = cur * 10 + t.charAt(a) - '0';
            }
            return n ? -cur : cur;
        }

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

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


}