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.io.*;
import java.util.*;
 
public class Main{
    public static void main(String[] args){ new Main().solve();}
    
    static Scanner scan = new Scanner(System.in);
    static final int MOD = 1_000_000_007;
    static final int INF = 2147483647;
 
    // Scanner
    int ni(){return Integer.parseInt(scan.next());}
    int[] ni(int n){int[] a = new int[n]; for(int i = 0; i < n; i++){a[i] = ni();} return a;}
    int[][] ni(int y, int x){int[][] a = new int[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ni();}} return a;}
    long nl(){return Long.parseLong(scan.next());}
    long[] nl(int n){long[] a = new long[n]; for(int i = 0; i < n; i++){a[i] = nl();} return a;}
    long[][] nl(int y, int x){long[][] a = new long[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = nl();}} return a;}
    String ns(){return scan.next();}
    String[] ns(int n){String[] a = new String[n]; for(int i = 0; i < n; i++){a[i] = ns();} return a;}
    String[][] ns(int y, int x){String[][] a = new String[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ns();}} return a;}
 
    // Math
    int max(int a, int b){return Math.max(a, b);}
    long max(long a, long b){return Math.max(a, b);}
    double max(double a, double b){return Math.max(a, b);}
    int max(int[] a){int max = a[0]; for(int value:a){max = max(max,value);} return max;}
    long max(long[] a){long max = a[0]; for(long value:a){max = max(max,value);} return max;}
    double max(double[] a){double max = a[0]; for(double value:a){max = max(max,value);} return max;}
    int min(int a, int b){return Math.min(a, b);}
    long min(long a, long b){return Math.min(a, b);}
    double min(double a, double b){return Math.min(a, b);}
    int min(int[] a){int min = a[0]; for(int value:a){min = min(min,value);} return min;}
    long min(long[] a){long min = a[0]; for(long value:a){min = min(min,value);} return min;}
    double min(double[] a){double min = a[0]; for(double value:a){min = min(min,value);} return min;}
    long sum(int[] a){long sum = 0; for(int value:a){sum += value;} return sum;}
    long sum(long[] a){long sum = 0; for(long value:a){sum += value;} return sum;}
    double sum(double[] a){double sum = 0; for(double value:a){sum += value;} return sum;}
    int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}
    long gcd(long a, long b){return b == 0 ? a : gcd(b, a % b);}
    int lcm(int a, int b){return a / gcd(a, b) * b;}
    long lcm(long a, long b){return a / gcd(a, b) * b;}
 
    // Array
    void sort(int[] a){ Arrays.sort(a);}
    void sort(long[] a){ Arrays.sort(a);}
    void sort(double[] a){ Arrays.sort(a);}
    void sort(String[] a){ Arrays.sort(a);}
    int[] reverse(int[] a){
        int[] reversed = new int[a.length];
        for(int i = 0; i < a.length; i++){
            reversed[a.length - i - 1] = a[i];
        }
        return reversed;
    }
    long[] reverse(long[] a){
        long[] reversed = new long[a.length];
        for(int i = 0; i < a.length; i++){
            reversed[a.length - i - 1] = a[i];
        }
        return reversed;
    }
    double[] reverse(double[] a){
        double[] reversed = new double[a.length];
        for(int i = 0; i < a.length; i++){
            reversed[a.length - i - 1] = a[i];
        }
        return reversed;
    }
    String[] reverse(String[] a){
        String[] reversed = new String[a.length];
        for(int i = 0; i < a.length; i++){
            reversed[a.length - i - 1] = a[i];
        }
        return reversed;
    }
    boolean[] reverse(boolean[] a){
        boolean[] reversed = new boolean[a.length];
        for(int i = 0; i < a.length; i++){
            reversed[a.length - i - 1] = a[i];
        }
        return reversed;
    }
    void fill(int[] array, int x) { Arrays.fill(array, x); }
    void fill(long[] array, long x) { Arrays.fill(array, x); }
    void fill(double[] array, double x) { Arrays.fill(array, x); }
    void fill(boolean[] array, boolean x) { Arrays.fill(array, x); }
    void fill(int[][] array, int x) { for(int a[] : array) { fill(a, x); } }
    void fill(long[][] array, long x) { for(long a[] : array) { fill(a, x); } }
    void fill(double[][] array, double x) { for(double a[] : array) { fill(a, x); } }
    void fill(boolean[][] array, boolean x) { for(boolean a[] : array) { fill(a, x); } }
 
 
    public void solve(){
		int n = ni();
		String[] a = ns(n);
		
		int cnt = 0;
		int aend = 0;
		int bstart = 0;
		int repeat = 0;
		for(int i = 0; i < n; i++){
			int tmp = 0;
			while(true){
				int index = a[i].indexOf("AB", tmp);
				if(index >= 0){
					cnt++;
					tmp = index + 2;
				} else {
					break;
				}
			}
			if(a[i].endsWith("A")){
				aend++;
			}
			if(a[i].startsWith("B")){
				bstart++;
			}
			if(a[i].endsWith("A") && a[i].startsWith("B")){
				repeat++;
			}
		}
		
		System.out.println(min(aend, bstart) == repeat ? cnt + min(aend, bstart) - 1 : cnt + min(aend, bstart));
    }
}