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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int abcount(string str){
  int c=0;
  while(str.find("AB")!=-1){
    c++;
    str.replace(str.find("AB"),2,"");
  }
  return c;
}

int main(){
  int N; cin >> N;
  string s[10000];
  for(int i=0; i<N; i++) cin >> s[i];
  
  //もともとあったABの個数
  long long result=0;
  for(int i=0; i<N; i++){
    result+=abcount(s[i]);
  }
  
  //並べたときに--A+B--となった個数
  int cc[4]={};
  for(int i=0; i<N; i++){
    if(s[i][0]=='B'){
      if(s[i][s[i].size()-1]=='A'){
        cc[3]++;
      }else{
        cc[1]++;
      }
    }else{
      if(s[i][s[i].size()-1]=='A'){
        cc[2]++;
      }else{
        cc[0]++;
      }
    }
  }
  
  result+=(cc[3]-1);
  if(cc[2]>0){
    result++;
    cc[2]--;
  }
  if(cc[1]>0){
    result++;
    cc[1]--;
  }
  result+=min(cc[1],cc[2]);
  
  cout << result << endl;
}