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 (GCC 5.4.1) (judge verdict: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); bool s=false; if(c=='-'){s=true;c=getchar_unlocked();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); int s=0; if(c=='-'){s=1;c=getchar_unlocked();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar_unlocked(); while(c==' '||c=='\n') c=getchar_unlocked(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar_unlocked(); } *s='\0'; return len; }

char s[11];
int main(void){
	int n=nextint();
	int a=0, b=0, c=0, d=0;
	for(int i=0; i<n; i++){
		int l=nextstr(s);
		a+=(s[0]=='B');
		b+=(s[l-1]=='A');
		d+=(s[0]=='B'&&s[l-1]=='A');
		for(int j=0; j<l-1; j++){
			if(s[j]=='A' && s[j+1]=='B') c++;
		}
	}
	int ans=c;
	if(a>b){
		ans+=b;
	}else if(a<b){
		ans+=a;
	}else{
		if(a>0){
			ans+=a-(a==d);
		}
	}
	printf("%d\n", ans);
}