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 <bits/stdc++.h>
using namespace std;
#define FOR(i,o,n) for(long long i = o;i<n;i++)
#define oneforall ios::sync_with_stdio(false);cin.tie(0);
#define all(v) (v).begin(),(v).end()
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define int long long 
const long long INF=1e18;
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
typedef vector<int> vi;
typedef vector<long long> vl;
typedef long long ll;
typedef vector<pair<long, long > > vpll;
typedef vector<pair<int, int > > vpii;



int32_t main() {
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	

	
	ini(n);
	int tmp = 0;
	int morty[3];
	FOR(i,0,n){
		ins(x);
		if(x[0] == 'B' and x[x.size()-1] == 'A')morty[0]++;
		else if(x[0] == 'B')morty[1]++;
		else if(x[x.size()-1] == 'A')morty[2]++;

		FOR(i1,0,x.size()-1){
			if(x[i1] == 'A' and x[i1+1] == 'B')tmp++;
		}
	}
	if(morty[1] ==0 and morty[2] == 0){
		tmp+=max(0ll,morty[0]-1);
	}
	else{
		tmp+=min(morty[1],morty[2]);
		tmp+=morty[0];
	}
	out(tmp);







	return 0;
}