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.*;
import java.util.*;

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args) {
        int N = in.i();
        char[] s = new char[10];

        int ans = 0;
        int A = 0;
        int B = 0;
        int AB = 0;
        for (int i = 0; i < N; i++) {
            int n = in.read(s);

            boolean readA = false;
            for (int j = 0; j < n; j++) {
                if (readA && s[j] == 'B') ans++;
                readA = s[j] == 'A';
            }

            boolean tailA = s[n - 1] == 'A';
            boolean headB = s[0] == 'B';
            if (tailA && headB) AB++;
            else if (tailA) A++;
            else if (headB) B++;
        }

        if (AB > 0) {
            ans += AB - 1;
            if (A > 0 || B > 0) ans++;
        }
        ans += Math.min(A, B);
        System.out.println(ans);
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = super.read(cbuf, 0, cbuf.length);
            } catch (IOException e) {
                e.printStackTrace();
            }
            head = 0;
        }
        return cbuf[head++];
    }

    int i() {
        int k = next() - '0';
        boolean minus = k == -3;
        int n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        int k = next() - '0';
        boolean minus = k == -3;
        long n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    char[] s(final int N) {
        char[] s = new char[N];
        for (int i = 0; i < N; i++) {
            s[i] = next();
        }
        next();
        return s;
    }

    public int read(char[] cbuf) {
        int i;
        char c;
        for (i = 0; (c = next()) != ' ' && c != '\n'; i++) cbuf[i] = c;
        return i;
    }

    public String readLine() {
        try {
            return super.readLine();
        } catch (IOException e) {
            return null;
        }
    }
}
