Codeforces Round #306 (Div. 2)——A——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.

Sample test(s)

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".

 智商被压制。。被hack了，只要倒着再判断一下就行

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
    char a[100110];
    while(~scanf("%s",&a)){
        int n = strlen(a);
        int t1 ,t2 ;
        t1 = t2 = 0;
        int i = 0;
        while(i < n){
            if(a[i] == 'A' && a[i+1] == 'B'&&t1 == 0){
            t1++;
            i++;}
            else if(a[i] == 'B' && a[i+1] == 'A'&&t2 == 0){
            t2++;
            i++;
            }
            i++;
        }
        if(t1 > 0 && t2 > 0){
            printf("YES\n");
            continue;
        }
         i = n-1;
        t1 = t2 = 0;
        while(i >= 0){
            if(a[i] == 'A' && a[i+1] == 'B'&& t1 == 0){
                t1++;
                i--;
            }
            else if(a[i] == 'B' && a[i+1] == 'A' && t2 == 0){
                t2++;
                i--;
            }
            i--;
        }
        if(t1 > 0 && t2 > 0)
        printf("YES\n");
        else printf("NO\n");
        }
    return 0;
}