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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

// https://atcoder.jp/contests/diverta2019/tasks/diverta2019_c
/*<head>*/
// #include "lib/Template.hpp"
/*</head>*/

/*<body>*/
/* #region header */
/**
 * @file Template.hpp
 * @brief 競技プログラミング用テンプレート
 * @author btk15049
 * @date 2019/05/02
 */
#include <bits/stdc++.h>
using namespace std;

/* #region macro */
#ifdef BTK
#    define DEBUG if (1)
#    define CIN_ONLY if (0)
#else
#    define DEBUG if (0)
#    define CIN_ONLY if (1)
#endif
/** @def
 * ALLマクロ
 */
#define ALL(v) (v).begin(), (v).end()

/** @def
 * 再帰ラムダをするためのマクロ
 */
#define REC(ret, ...) std::function<ret(__VA_ARGS__)>
/* #endregion */

namespace _Template_ {
    /**
     * @brief cin高速化処理を行うための構造体
     * @details CIN_ONLYマクロで動作が変わる
     */
    struct cww {
        cww() {
            CIN_ONLY {
                ios::sync_with_stdio(false);
                cin.tie(0);
            }
        }
    } star;
    /**
     * @brief change min
     * @tparam T 型
     * @param l 参照
     * @param r 非参照
     * @return 更新があればtrue
     */
    template <typename T>
    inline bool chmin(T& l, T r) {
        bool a = l > r;
        if (a) l = r;
        return a;
    }
    /**
     * @brief chminのmax版
     * @see chmin
     */
    template <typename T>
    inline bool chmax(T& l, T r) {
        bool a = l < r;
        if (a) l = r;
        return a;
    }
    /**
     * @brief
     * vectorに直接cin流すためのやつ
     * @tparam T
     * @param is
     * @param v
     * @return istream&
     */
    template <typename T>
    istream& operator>>(istream& is, vector<T>& v) {
        for (auto& it : v) is >> it;
        return is;
    }
    /**
     * @brief
     * rangeを逆向きに操作したいとき用
     * @details
     * ループの範囲は[bg,ed)なので注意
     * @see range
     */
    class reverse_range {
      private:
        struct I {
            int x;
            int operator*() { return x - 1; }
            bool operator!=(I& lhs) { return x > lhs.x; }
            void operator++() { --x; }
        };
        I i, n;

      public:
        reverse_range(int n) : i({0}), n({n}) {}
        reverse_range(int i, int n) : i({i}), n({n}) {}
        I& begin() { return n; }
        I& end() { return i; }
    };
    /**
     * @brief
     * python みたいな range-based for を実現
     * @details
     * ループの範囲は[bg,ed)なので注意
     * !つけると逆向きにループが回る (reverse_range)
     * 空間計算量はO(1)
     * 使わない変数ができて警告が出がちなので，unused_varとかを使って警告消しするとよい
     */
    class range {
      private:
        struct I {
            int x;
            int operator*() { return x; }
            bool operator!=(I& lhs) { return x < lhs.x; }
            void operator++() { ++x; }
        };
        I i, n;

      public:
        range(int n) : i({0}), n({n}) {}
        range(int i, int n) : i({i}), n({n}) {}
        I& begin() { return i; }
        I& end() { return n; }
        reverse_range operator!() { return reverse_range(*i, *n); }
    };
    /**
     * @brief
     * rangeで生まれる使わない変数を消す用（警告消し）
     */
    template <typename T>
    inline T& unused_var(T& v) {
        return v;
    }
    using LL = long long;
} // namespace _Template_
using namespace _Template_;

/* #endregion */
/*</body>*/
constexpr int A = 1;

constexpr int B  = 2;
constexpr int BA = 3;
int main() {
    int N;
    cin >> N;
    vector<int> cnt(4);
    int ret = 0;
    for (int i : range(N)) {
        string s;
        cin >> s;
        int m = 0;
        if (s.front() == 'B') m |= 1;
        if (s.back() == 'A') m |= 2;
        cnt[m]++;
        for (int si : range(1, s.size())) {
            if (s[si - 1] == 'A' && s[si] == 'B') {
                ret++;
            }
        }
    }
    if (cnt[BA]) {
        ret += cnt[BA] - 1;
        cnt[BA] = 1;
    }
    if (cnt[A] && cnt[BA]) {
        ret++;
        cnt[A]--;
    }
    cnt[A] += cnt[BA];
    ret += min(cnt[A], cnt[B]);
    cout << ret << endl;
    return 0;
}