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: Accepted)

/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
'+'とかは問題ない
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
負の数の割り算は絶対値が小さいほうに切り捨てられるex.-1/2=0ので、ある値>kみたいな式だとバグらせやすいので注意
ある値>=kとして、切り上げた値をとるべき
toLowerCase()はアルファベット以外に対して行ってもエラーにはならない
ArrayDequeではpopとpushはStackと同じ操作になる
//import static java.lang.System.*; はしばらく使わなそう(使うとしてもexit(0)のためだけ)←return;でよさそう
最後にout.flush()またはout.close()をするのを忘れずに！
Point2D p2d=new Point2D.Double(d1,d2);
next_Permutationを使うときはソート済み配列かどうか確認！！
*/
//入力終了→Ctrl+D
import java.util.*;
import java.io.*;
import java.awt.*;
import java.awt.geom.Point2D;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        int n=sc.nextInt();
        int a=0,b=0,c=0,ab=0;
        for (int i = 0; i < n; i++) {
            String s = sc.next();
            int len = s.length();
            for (int j = 1; j < len; j++) {
                if (s.charAt(j - 1) == 'A' && s.charAt(j) == 'B') ab++;
            }
            if (s.charAt(0) == 'B' && s.charAt(len - 1) == 'A') c++;
            else if (s.charAt(len - 1) == 'A') a++;
            else if (s.charAt(0) == 'B') b++;
        }
        out.println(ab+(c==0?min(a,b):(a+b==0?c-1:c+min(a,b))));
        out.close();
    }

    //aはソート済みでなければならない！
    static boolean nextPermutation(int[] a) {
        int t = a.length - 1;
        int tt = t + 1;
        for (int i = t; i > 0; --i) {
            if (a[i - 1] < a[i]) {
                int swapIndex = Find(a[i - 1], a, i, t);
                int temp = a[swapIndex];
                a[swapIndex] = a[i - 1];
                a[i - 1] = temp;
                Arrays.sort(a, i, tt);
                return true;
            }
        }
        return false;
    }

    static int Find(int dest, int[] a, int s, int e) {
        if (s == e) {
            return s;
        }
        int m = (s + e + 1) / 2;
        return a[m] <= dest ? Find(dest, a, s, m - 1) : Find(dest, a, m, e);
    }

    static long power(long x, int n) {
        if (n == 0) return 1;
        if (n % 2 == 0) {
            long e = power(x, n / 2);
            return (e * e) % mod;
        }
        return (x * power(x, n - 1)) % mod;
    }

    static long mod = (long) 1e9 + 7;

    //クラスカル法(0-indexed) n頂点m辺
    static HashSet<Integer> Kruskal(int[] u, int[] v, long[] cost, int n) {
        int m = u.length;
        UF.Init(n);
        PriorityQueue<Integer> S = new PriorityQueue<>((p1, p2) -> Long.compare(cost[p1], cost[p2]));
        for (int i = 0; i < m; i++) {
            S.add(i);
        }
        //最小全域木をなす辺
        HashSet<Integer> F = new HashSet<>();
        while (!S.isEmpty()) {
            int e = S.poll();
            if (!UF.same(u[e], v[e])) {
                F.add(e);
                UF.unite(u[e], v[e]);
            }
        }
        return F;
    }

    //ワーシャルフロイド(0-indexed)
    static void WF(long[][] d, int[] a, int[] b, long[] c) {
        int n = d[0].length;
        long INF = 100000000000l;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                d[i][j] = d[j][i] = INF;
            }
        }
        int m = a.length;
        for (int i = 0; i < m; i++) {
            d[a[i]][b[i]] = d[b[i]][a[i]] = c[i];
        }
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    d[i][j] = d[i][j] > d[i][k] + d[k][j] ? d[i][k] + d[k][j] : d[i][j];
                }
            }
        }
    }

    //Union Find木(0-indexed)
    static class UF {
        static int[] d;

        static void Init(int size) {
            d = new int[size];
            for (int i = 0; i < size; i++) {
                d[i] = -1;
            }
        }

        static boolean unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return false;
            if (d[x] > d[y]) {
                int t = d[y];
                d[y] = d[x];
                d[x] = t;
            }
            d[x] += d[y];
            d[y] = x;
            return true;
        }

        static boolean same(int x, int y) {
            return find(x) == find(y);
        }

        static int find(int k) {
            return d[k] < 0 ? k : (d[k] = find(d[k]));
        }

        static int size(int k) {
            return -d[find(k)];
        }
    }

    static long gcd(long a, long b) {
        return a % b == 0 ? b : gcd(b, a % b);
    }

    static PrintWriter out = new PrintWriter(System.out);

    static class sc {
        static Scanner s = new Scanner(System.in);

        static String next() {
            return s.next();
        }

        static String nextLine() {
            return s.nextLine();
        }

        static int nextInt() {
            return Integer.parseInt(s.next());
        }

        static long nextLong() {
            return Long.parseLong(s.next());
        }

        static double nextDouble() {
            return Double.parseDouble(s.next());
        }

        static boolean hasNext() {
            return s.hasNext();
        }

        static boolean hasNextInt() {
            return s.hasNextInt();
        }

    }

}