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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep0(i,n) for(ll (i) = 0; (i) < (n); ++(i))
#define rrep0(i,n) for(ll (i) = (n) - 1; (i) >= 0; --(i))
#define rep1(i,n) for(ll (i) = 1; (i) <= (n); ++(i))
#define rrep1(i,n) for(ll (i) = (n); (i) >= 1; --(i))
#define nfor(i,a,b) for(ll (i) = (a); (i) < (b); ++(i))
#define rnfor(i,a,b) for(ll (i) = (a) - 1; (i) >= (b); --(i))
#define ll long long
#define mod (1000000007)
#define pf(x) cout << (x) << endl
#define all(x) (x).begin(),(x).end()

ll gcd(ll a,ll b){
    if(a < b)swap(a, b);
    if(b == 0) return a;
    return gcd(b,a%b);
}

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}

ll factorial(ll n){
    ll t = 1;
    rep1(i, n)t *= i;
    return t;
}

ll combinations(ll n, ll k) {
    ll r = 1;
    rep1(i, k)r = r * n-- / i;
    return r;
}

int main(){
    ll n,ans = 0,cnta = 0,cntb = 0,cntab = 0;
    cin >> n;
    rep0(i, n){
        string s;
        cin >> s;
        if(s[0] == 'B' && s[s.size() - 1] == 'A')++cntab;
        if(s[s.size() - 1] == 'A')++cnta;
        if(s[0] == 'B')++cntb;
        rep0(j, s.size() - 1)if(s[j] == 'A' && s[j + 1] == 'B')++ans;
    }
    ans += min(cnta,cntb);
    if(cnta == cntb && cntab == cnta)--ans;
    pf(ans);
    return 0;
}