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)

#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const llint mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}

int main(void){
	int i,n;cin>>n;
	int ans=0,A=0,B=0,kata=0;
	for(i=0;i<n;i++){
		string str;cin>>str;
		for(int j=0;j+1<str.size();j++){
			if(str[j]=='A'&&str[j+1]=='B'){ans++;}
		}
		if(str.back()=='A'){A++;}
		if(str[0]=='B'){B++;}
		if((str.back()=='A')!=(str[0]=='B')){kata=1;}
	}
	cout<<ans+max(0,min(A,B)-1+kata)<<endl;
}