Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time. 

During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*. 

Let's help Vitaly solve this easy problem!
The first line contains string *s* (1 ≤ |*s*| ≤ 100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.

The second line contains string *t* (|*t*| = |*s*|), consisting of lowercase English letters.

It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
# y tuong:
# duyet mang tu dau den cuoi, tao ra mang khac mậc định toàn False
# xet tung ki tu trong chuoi s va t
# neu hieu s[i] va t[i] lon hơn bằng 2, danh dấu check[i] = True
#    Neu la True và t[i] khac ki tu 'a' thi đưa ra két quả là t[:i]+(t[i]-1)+t[i+1:]
#    Nguoc lai ket qua la t[:i]+(s[i]+1)+t[i+1:]
# Neu hiệu s[i] va t[i] bằng 1: cho j = i+1, lặp với j
#    Neu t[j] khác a: thi ket qua: t[:j] + 'a'
#    Neu s[j] bằng a thì: kết quả là: s[:j] + 'b'

s = input ()
t = input()
def string (string1, string2):
  ans = "No such string"
  check = [False]* len(string1)
  i = 0
  while i < len(string1):
    if abs(ord(string1[i]) - ord(string2[i])) >= 2 and string2[i]!= 'a':
      check [i] = True
    if True in check:
      ans = string2
      if string2[i] == 'a':
        ans = ans[:i]+chr(ord(string1[i])+1) + ans[i+1:]
      else:
        ans = string2[:i]+chr(ord(string2[i])-1) + ans[i+1:]
      return ans
    if ord(string2[i]) - ord(string1[i]) == 1:
      j = i+1
      while j < len(string1):
        if string2[j] != 'a' :
          ans =""
          ans = string2[:j] + 'a'
        if string1[j] == 'a':
          ans =""
          ans = string1[:j] + 'b'

        j += 1
    i += 1
  return ans

print(string(s,t))
