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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

"use strict";
function main(input) {
  const args = input.split("\n");

  const N = parseInt(args[0]);

  const p = /(AB)+/g;

  let testStr;
  let testStrLen;
  let countAB = 0;

  let BA = 0;
  let firstB = 0;
  let lastA = 0;

  for(let i = 0; i < N; i++) {
      testStr = args[i + 1];
      testStrLen = testStr.length;
      while(p.exec(testStr)) {
          countAB ++;
      }
      if (testStr.charAt(0) == "B") {
        if (testStr.charAt(testStrLen - 1) == "A") {
          BA++;
        } else {
          firstB++;
        }
      } else if (testStr.charAt(testStrLen - 1) == "A") {
        lastA++;
      }
  }


  countAB += BA - 1;
  if(firstB > 0) {
      countAB ++;
      firstB --;
  }

  if(lastA < firstB) {
      countAB += lastA;
  } else {
      countAB += firstB
  }

  console.log(countAB);
}
main(require("fs").readFileSync("/dev/stdin", "utf8"));