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

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 test(s)

input

a
c

output

b

input

aaa
zzz

output

kkk

input

abcdefg
abcdefh

output

No such string

Note

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.

大意：判断是否存在处于两个字符串之间的串，第二个串已经加了一，那么从最后开始加1若果超过了26，那么前面那个就加1，比较之后的串是否相同，相同的话说明不存在。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
    char a[110],b[110];
    cin>>a>>b;
    int n = strlen(a);
    a[n-1]++;
    for(int i = n - 1; i >= 0 ;i--){
            if(a[i]>'z'){
                a[i] = 'a';
                a[i-1]++;
            }
    }
    if(strcmp(a,b) == 0)
        printf("No such string
");
        else {
                printf("%s",a);
                printf("
");
       }
    return 0;
}