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: Wrong Answer)

#include<stdio.h>

int len(char s[10]){
    int length = 0;
    while(s[length])    {
        length ++;
    }
    return length;
}

long long max (long long a, long long b)
{
    if(a>b) return a;
    else return b;
}
int main(void)
{
    long long N;
    scanf("%lld",&N);
    char s[N][10];
    int b[N];
    long long i;
    for (i = 0; i < N; i ++)    {
        scanf("%s",s[i]);
        if(i != N-1)    scanf("%d",&b[i]);
    }   //入力テスト済み//
    long long ans;
    ans = 0;
    for (i = 0; i < N; i++) {
        for (int j = 0; j < 9; j ++)   {
            if(s[i][j] == 'A' && s[i][j+1] =='B')   ans ++;
        }
    }
    long long lasta = 0;
    long long firstb = 0;
    long long both = 0;
    for (i = 0; i < N; i++) {
        if(s[i][0] == 'B')  firstb ++;
        if(s[i][len(s[i])-1] == 'A')  lasta ++; 
        if(s[i][0] == 'B' && s[i][len(s[i])-1] == 'A')    both ++;
    }
    firstb -= both;
    lasta -= both;
    if(firstb > lasta)  {
        long long temp;
        temp = firstb;
        firstb = lasta;
        lasta = temp;
    } //make lasta bigger than firstb//
    if(lasta - firstb <= both)  {
        ans += firstb + both;
    }
    else
    {
        ans += lasta + (both - (lasta- firstb))/2;
    }
    
    printf("%lld",ans); 
}