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 <iostream>
#include <stack>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <functional>
#include <string>
#include <cstdlib>
#include <numeric>
#include <cstdbool>
#include <map> 
#include<set>
#include<queue>

typedef long long ll;


#define rep(i,n) for(ll i=0;i<n;i++)
using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long INF = 1000000000000000;
const ll inf = -1e18;
typedef pair<int, int> P;
ll ma = 1000000000 + 7;
ll h, w, n,k,m; string s,t;
char maze[60][60];
int dis[101][101]; int graph[50][50]; bool vis[50];
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };
ll gcd(ll x, ll y) {
	if (x % y == 0) return y;
	return gcd(y, x % y);

}
ll lcm(ll a,ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
int main() {
	cin >> n;
	int res = 0;
	int bcnt = 0; int acnt = 0; int cntab = 0;
	rep(i, n) {
		string str;
		cin >> str;
		rep(j, str.size()-1) {
			if (str[j] == 'A') {
				if (str[j + 1] == 'B') {
					res++;
				}
			}
		}
		bool flg = true;
		if (str[0] == 'B') {
			bcnt++; flg = false;
		}
		int f = str.size() - 1;
		if (str[f] == 'A') {
			acnt++;
			if (!flg) {
				acnt--; bcnt--;
				cntab++;
			}
		}
	}
	res += min(bcnt, acnt);
	res += cntab - 1;
	if (abs(acnt - bcnt) > 0&&cntab>0) {
		res++;
	}
	cout << res << endl;
}