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.
# print("Input the string")
st = input()

if st == 'a':
    print('z')
else:
    apos = st.find('a')
    if apos == -1:
        #  No as in the word at all--shift everything down
        answerlist = []
        for ch in st:
            newch = chr(ord(ch)-1)
            answerlist.append(newch)
        print(''.join(answerlist))
    elif apos > 0:
        #  Shift everything up to the first a
        answerlist = []
        for i in range(0, apos):
            newch = chr(ord(st[i])-1)
            answerlist.append(newch)
        print(''.join(answerlist) + st[apos:])
    else:
        #  The hard case--it starts with an a
        #  Find the first non-a
        pos = 1
        while pos < len(st) and st[pos] == 'a':
            pos += 1
        if pos == len(st):  # All letters are a--must change last letter to z
            print(st[0:len(st)-1] + 'z')
        else:
            # Change everything from pos to the next a
            nextapos = st.find('a', pos+1)
            if nextapos == -1:  # No more as--change to the end
                answerlist = []
                for i in range(pos, len(st)):
                    newch = chr(ord(st[i])-1)
                    answerlist.append(newch)
                print(st[0:pos] + ''.join(answerlist))
            else:   # Change everything between pos and nextapos
                answerlist = []
                for i in range(pos, nextapos):
                    newch = chr(ord(st[i])-1)
                    answerlist.append(newch)
                print(st[0:pos] + ''.join(answerlist) + st[nextapos:])
             
            
        
        
