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.

Examples

Input

3
ABCA
XBAZ
BAD


Output

2


Input

9
BEWPVCRWH
ZZNQYIJX
BAVREA
PA
HJMYITEOX
BCJHMRMNK
BP
QVFABZ
PRGKSPUNA


Output

4


Input

7
RABYBBE
JOZ
BMHQUVA
BPA
ISU
MCMABAOBHZ
SZMEHMA


Output

4
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    private static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {

        int n = sc.nextInt();

        List<String> s = new ArrayList<>();
        int endsWithA = 0;
        int beginsWithB = 0;
        int both = 0;
        for (int i=0; i < n; i++) {
            String t = sc.next();
            s.add(t);
            if (t.startsWith("B") && !t.endsWith("A"))
                beginsWithB++;
            else if (t.endsWith("A") && !t.startsWith("B"))
                endsWithA++;
            else if (t.endsWith("A") && t.startsWith("B"))
                both++;
        }

        int ans = 0;
        for (String e : s) {
            ans += containsHowManyAB(e);
        }

        if (both > 1) {
            ans += both - 1;
        }
        if (both > 0) {
            if (endsWithA > 0) {
                endsWithA--;
                ans++;
            }
            if (beginsWithB > 0) {
                beginsWithB--;
                ans++;
            }
        }

        ans += endsWithA > beginsWithB ? beginsWithB : endsWithA;
        System.out.println(ans);
    }

    private static int containsHowManyAB(String s) {
        int ret = 0;
        for (int i=0; i < s.length()-1; i++) {
            if (s.substring(i, i+2).equals("AB"))
                ret++;
        }
        return ret;
    }

}
