Equivalent Strings

Description

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are calledequivalent in one of the two cases:

1. They are equal.
2. If we split string a into two halves of the same size a₁ and a₂, and string b into two halves of the same size b₁ and b₂, then one of the following is correct:
   1. a₁ is equivalent to b₁, and a₂ is equivalent to b₂
   2. a₁ is equivalent to b₂, and a₂ is equivalent to b₁

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it's your turn!

Input

The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.

Output

Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample Input

Input

aaba
abaa

Output

YES

Input

aabb
abab

Output

NO

Hint

In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".

 

这道题使用递归枚举，不过要注意剪枝，不然会Memory limit exceeded，原因是堆栈开的太多了。你要一直递归到字符串分为1，那实在是递归太多了.

可以使用string类做，也可以使用char数组加指针，前者的空间开销和时间开销都要大一些

 

方法一  string类：

#include"iostream"
using namespace std;

int judge(string a,string b)
{
     if(a==b) return 1;
     if(a.length()%2) return 0;
     if(a.length()==1&&b.length()==1)
     {
   //      cout<<a[0]<<"    "<<b[0]<<endl;
     if(a[0]==b[0])
     return 1;
     else return 0;
     }
     string a1,a2,b1,b2;
     a1=a.substr(0,a.length()/2);
     a2=a.substr(a.length()/2,a.length());
     b1=b.substr(0,b.length()/2);
     b2=b.substr(b.length()/2,b.length());if(judge(a1,b2)==1&&judge(a2,b1)==1) return 1;
     if(judge(a1,b1)==1&&judge(a2,b2)==1)  return 1;

      return 0;
}
int main()
{
    string a,b,a1,a2;
    while(cin>>a>>b)
    {
        if(judge(a,b)) cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}

 

方法二   指针

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;

const int maxn=2700010;
char a[maxn],b[maxn];

int judge(char* a,char* b,int len)
{
   if(!strncmp(a,b,len)) return 1;
   if(len%2) return 0;
     
       if((judge(a,b+len/2,len/2)==1&&judge(a+len/2,b,len/2)==1)) {return 1;}
       if((judge(a,b,len/2)==1&&judge(a+len/2,b+len/2,len/2)==1)) return 1;
      return 0;
}
int main()
{
       /* gets(a);
        getchar();
        gets(b);*/
        cin>>a>>b;
        if(judge(a,b,strlen(a))) cout<<"YES\n";
        else cout<<"NO\n";
    return 0;
}