Problem:
Score : 400 points

Problem Statement

Snuke has N strings. The i-th string is s_i.

Let us concatenate these strings into one string after arranging them in some order.
Find the maximum possible number of occurrences of AB in the resulting string.

Constraints

1 \leq N \leq 10^{4}

2 \leq |s_i| \leq 10

s_i consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
s_1
\vdots
s_N

Output

Print the answer.

Sample Input 1

3
ABCA
XBAZ
BAD

Sample Output 1

2

For example, if we concatenate ABCA, BAD and XBAZ in this order, the resulting string ABCABADXBAZ has two occurrences of AB.

Sample Input 2

9
BEWPVCRWH
ZZNQYIJX
BAVREA
PA
HJMYITEOX
BCJHMRMNK
BP
QVFABZ
PRGKSPUNA

Sample Output 2

4

Sample Input 3

7
RABYBBE
JOZ
BMHQUVA
BPA
ISU
MCMABAOBHZ
SZMEHMA

Sample Output 3

4

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.Map.Entry;
 
class Main {
	 static int mod =  (int) (Math.pow(10,9)+7);
//	 static int mod = 998244353;
   public static void main(String[] args) {
	   
       Scanner sc = new Scanner(System.in);
       int N = sc.nextInt();
//       int B = sc.nextInt();
//       String S = sc.next();
//       String T = sc.next();
//       char[] s = S.toCharArray();
//       char[] t = T.toCharArray();
       int f = 0;
       int l = 0;
       int fl = 0;
       int cnt=0;
       for (int i=0 ; i<N; i++) {
           String S = sc.next();
           char[] s = S.toCharArray();
           if(s[0]=='B'&& s[s.length-1]=='A') {
        	   fl++;
           } else if(s[0]=='B') {
        	   f++;
           } else if(s[s.length-1]=='A') {
        	   l++;
           }
           while(S.indexOf("AB")!=-1) {
        	   S=S.substring(S.indexOf("AB")+2);
        	   cnt++;
           }
       }
       
	   int k1 =Math.min(fl, f);
	   int k2 =Math.min(fl,l);
	   int k3 =Math.min(f,l);
	   cnt=cnt+k1;
	   cnt=cnt+k2;
	   f=f-k1;
	   l=l-k2;
	   fl=fl-k3;
	   
       if(fl>0) {
    	   cnt=cnt+fl-1;
       }
       cnt=cnt+Math.min(f, l);
       
       System.out.println(cnt);
   }
      

}