Two Substrings

You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings “AB” and “BA” (the substrings can go in any order).

Input

The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.

Output

Print “YES” (without the quotes), if string s contains two non-overlapping substrings “AB” and “BA”, and “NO” otherwise.

Examples

Input
ABA
Output
NO
Input
BACFAB
Output
YES
Input
AXBYBXA
Output
NO

Note

In the first sample test, despite the fact that there are substrings “AB” and “BA”, their occurrences overlap, so the answer is “NO”.

In the second sample test there are the following occurrences of the substrings: BACFAB.

In the third sample test there is no substring “AB” nor substring “BA”.

Note

In the first sample test, despite the fact that there are substrings “AB” and “BA”, their occurrences overlap, so the answer is “NO”.

In the second sample test there are the following occurrences of the substrings: BACFAB.

In the third sample test there is no substring “AB” nor substring “BA”.

题目大意：

在字符串中是否存在一个“AB”和一个“BA”，这两个串不交叉。

解题思路：

模拟即可。。。代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
  string s;
  cin>>s;
  bool flag1=false,flag2=false;
  for(int i=0;i<s.size();i++)
  {
    if(s[i]=='A'&&s[i+1]=='B')
    {
      if(!flag1)
      for(int j=i+2;j<s.size();j++)
      {
        if(s[j]=='B'&&s[j+1]=='A')
        {
          cout<<"YES"<<endl;
          return 0;
        }
      }
      if(!flag1) flag1=true;
    }
    else if(s[i]=='B'&&s[i+1]=='A')
    {
      if(!flag2)
      for(int j=i+2;j<s.size();j++)
      {
        if(s[j]=='A'&&s[j+1]=='B')
        {
          cout<<"YES"<<endl;
          return 0;
        }
      }
      if(!flag2)flag2=true;
    }
    if(flag1&&flag2)break;
  }
  cout<<"NO"<<endl;
  return 0;
}