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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
 
public class Main {
 
    void solve() throws IOException {
        int N = ni();
        int sb = 0;
        int ea = 0;
        int sbea = 0;
        int ans = 0;
        for (int i = 0; i < N; i++) {
            String s = ns();
            if(s.startsWith("B") && (!s.endsWith("A"))) sb++;
            if(s.endsWith("A") && (!s.startsWith("B"))) ea++;
            if(s.startsWith("B")&&s.endsWith("A")) sbea++;
            for (int j = 0; j < s.length()-1; j++) {
                if(s.substring(j,j+2).equals("AB")) ans++;
            }
        }
        if(sbea==0){
            if(sb+ea>1)ans += Math.min(sb,ea);
        }else{
            if(sb==0 && ea==0) ans += sbea-1;
            else ans += sbea + Math.min(sb,ea);
        }
        out.println(ans);
    }

    final int mod = 1000000007;
    int mul(int x, int y){
        return (int)((x * 1L * y) % mod);
    }
    int add(int x, int y) {
        x += y;
        return x >= mod ? (x - mod) : x;
    }
    int sub(int x, int y){
        x = add(x,mod-y);
        return x >= mod ? (x - mod) : x;
    }
    String ns() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine(), " ");
        }
        return tok.nextToken();
    }
 
    int ni() throws IOException {
        return Integer.parseInt(ns());
    }
 
    long nl() throws IOException {
        return Long.parseLong(ns());
    }
 
    double nd() throws IOException {
        return Double.parseDouble(ns());
    }
 
    String[] nsa(int n) throws IOException {
        String[] res = new String[n];
        for (int i = 0; i < n; i++) {
            res[i] = ns();
        }
        return res;
    }
 
    int[] nia(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = ni();
        }
        return res;
    }
 
    long[] nla(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nl();
        }
        return res;
    }
 
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
 
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        tok = new StringTokenizer("");
        Main main = new Main();
        main.solve();
        out.close();
    }
}