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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define REP(i, n) for(int i=0; i<(n); ++i)
#define FOR(i, a, b) for(int i=(a); i<(b); ++i)
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)

#define DEBUG(x) cout<<#x<<": "<<(x)<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";REP(i, v.size())cout<<' '<<v[i];cout<<'\n'
#define ALL(a) (a).begin(), (a).end()

template<typename T> inline void CHMAX(T& a, const T b) {if(a<b) a=b;}
template<typename T> inline void CHMIN(T& a, const T b) {if(a>b) a=b;}

constexpr ll MOD=1000000007ll;
// constexpr ll MOD=998244353ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

const double EPS=1e-11;
#define EQ0(x) (abs((x))<EPS)
#define EQ(a, b) (abs((a)-(b))<EPS)

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	// cout<<setprecision(10)<<fixed;
	int n;
	string s[12525];
	cin>>n;
	int a=0, b=0, ans=0, ba=0;
	REP(i, n){
		cin>>s[i];
		if(s[i][0]=='B'){
			++b;
		}
		int sz=s[i].size();
		if(s[i][sz-1]=='A'){
			++a;
		}
		if(s[i][0]=='B' && s[i][sz-1]=='A'){
			++ba;
		}
		bool flag=false;
		REP(j, sz){
			if(s[i][j]=='A'){
				flag=true;
			}
			if(flag && s[i][j]=='B'){
				++ans;
				flag=false;
			}
			if(s[i][j]!='A'){
				flag=false;
			}
		}
	}
	if(a==ba && b==ba){
		ans+=max(0, min(a, b)-1);
	}
	else{
		ans+=min(a, b);
	}
	cout<<ans<<'\n';
	return 0;
}