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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <vector>
#include <string.h>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <math.h>
#include <algorithm>
#include <numeric>

using namespace std;

// マクロ&定数 ============================================================
typedef unsigned int uint;
typedef long long ll;
//typedef pair<int, int> P;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<bool>> vvbool;

const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = { 0,0,1,-1, 1, 1, -1, -1, 0 }; // 4;４近傍
const int DY[9] = { 1,-1,0,0, 1, -1, 1, -1, 0 }; // 8:８近傍 9:(0,0)を含む
const ll MOD = 1e9 + 7; //10^9 + 7
const double PI = 3.14159265358979323846264338327950288;
//========================================================================


int main() {

	//==================================
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(30);
	//==================================

	ll N;
	cin >> N;

	vstring S(N);
	for (ll i = 0; i < N; i++) {
		cin >> S[i];
	}

	ll middle = 0;
	ll start_B = 0;
	ll end_A = 0;

	for (ll i = 0; i < N; i++) {
		if (S[i][0] == 'B') {
			start_B++;
		}
		
		if (S[i][S[i].length() - 1] == 'A') {
			end_A++;
		}

		for (ll j = 0; j < S[i].length() - 1; j++) {
			if (S[i][j] == 'A' && S[i][j + 1] == 'B') {
				middle++;
			}
		}


	}


	cout << middle + min(min(start_B, end_A), N-1);


}