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 io, os
import sys
from atexit import register

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.stdout = io.BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))

tokens = []
tokens_next = 0


def nextStr():
    global tokens, tokens_next
    while tokens_next >= len(tokens):
        tokens = input().split()
        tokens_next = 0
    tokens_next += 1

    if type(tokens[tokens_next - 1]) == str:
        return tokens[tokens_next - 1]
    return tokens[tokens_next - 1].decode()


def nextInt():
    return int(nextStr())


def nextIntArr(n):
    return [nextInt() for i in range(n)]


def print(s, end='\n'):
    sys.stdout.write((str(s) + end).encode())


def shift(c):
    if c == 'a':
        return 'z'
    res = ord(c) - 1
    return chr(res)


s = nextStr()

nonA = next((i for i in range(len(s)) if s[i] != 'a'), len(s))
if nonA == len(s):
    print(s[:-1] + shift(s[-1]))
    exit(0)

nextA = next((i for i in range(nonA, len(s)) if s[i] == 'a'), len(s))
print(s[:nonA] + ''.join(shift(c) for c in s[nonA:nextA]) + s[nextA:])
