Codeforces Round #293 (Div. 2) -- A. Vitaly and Strings (字符串构造)

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本文介绍了一个字符串构造问题，任务是找到一个字符串u，该字符串在字典序上大于s且小于t。文章提供了实现思路和代码示例，展示了如何通过修改字符串s来生成所需的字符串u。

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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.

Examples

Input

a
c

Output

Input

aaa
zzz

Output

kkk

Input

abcdefg
abcdefh

Output

No such string

Note

String s = s₁s₂... s_n is said to be lexicographically smaller than t = t₁t₂... t_n, if there exists such i, that s₁ = t₁, s₂ = t₂, ... s_i - 1 = t_i - 1, s_i < t_i.

大体题意：

给你两个字符串s，t，构造一个字符串u 使得 u > s && u < t 如果不存在输出

No such string

思路：

直接对s字符串进行操作，给这个字符串加1，如果字符是‘z’ 就变成a

这样的字符串是最小的比s 大的字符串，如果他比t小就是他了，否则不存在！！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100 + 10;
const int INF = 0x3f3f3f3f;
char s1[maxn],s2[maxn],s[maxn];
int main(){
    while(scanf("%s%s",s1,s2) == 2){
        int len = strlen(s1);
        for (int i = len-1; i >= 0; --i){
            if (s1[i] != 'z'){
                s1[i]++;
                break;
            }
            else s1[i] = 'a';
        }
        if (strcmp(s1,s2) < 0)puts(s1);
        else puts("No such string");
    }
    return 0;
}