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

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.

题意：给你两个字符串a,b，然后让你输出一个字符串，这个字符串c要求满足 a<c<b

坑点在于z++之后，就不是字母了，这点需要判断一下就是了

string a;
string b;
string c;
int main()
{
        cin>>a>>b;
        c=a;
        for(int i=a.size()-1;i>=0;i--)
        {
                if(c[i]=='z')
                        c[i]-=25;
                else
                {
                        c[i]++;
                        break;
                }
        }
        int flag=0;
        for(int i=0;i<a.size();i++)
        {
                if(c[i]<b[i])
                {
                        flag=1;
                        break;
                }
                if(c[i]>b[i])
                {
                        flag=0;
                        break;
                }
        }
        if(flag==0)
                cout<<"No such string"<<endl;
        else
                cout<<c<<endl;
}