Codeforces Round #293 (Div. 2)--A-- Vitaly and Strings - Painting

小水，WA了几次，其实不用想太多，因为str1的字典序题目中说肯定比str2的小，所以只要让str1的字典序加上1即可，所以要倒着循环，如果最后是z就要考虑进位的因素。最后在比较一下，str1 ！= str2 就输出就行了！！！

Description

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!

Input

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.

Output

If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).

If such string exists, print it. If there are multiple valid strings, you may print any of them.

Sample Input

Input

a
c

Output

b

Input

aaa
zzz

Output

kkk

Input

abcdefg
abcdefh

Output

No such string

Hint

String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn,
if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstdlib>
#include<string>
#include<string.h>

using namespace std;

string str1;
string str2;

int main()
{
    cin>>str1>>str2;
    int len = str1.length();
    for(int i = len -1; i >= 0 ; i--)
    {
        if(str1[i] != 'z')
        {
            str1[i] = str1[i] + 1;
            break;
        }
        if(str1[i] == 'z')
            str1[i] = 'a';
    }
    if(str1 != str2)
        cout<<str1<<endl;
    else
        cout<<"No such string"<<endl;
    return 0;
}