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 9.2.1) (judge verdict: Accepted)

constexpr bool isDebugMode = false;
int testcase = 1;
#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
    #include <atcoder/all>
    using namespace atcoder;
#endif
using namespace std;
typedef long long ll;
typedef pair<long long, long long> P;
struct edge{long long to,cost;};
const int inf = 1 << 27;
const long long INF = 1LL << 60;
const int COMBMAX = 1001001;
const long long MOD = 1000000007;
const long long dy[] = {-1, 0, 0, 1};
const long long dx[] = {0, -1, 1, 0};
string abc = "abcdefghijklmnopqrstuvwxyz";
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for (const auto &e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (long long)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (long long)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
long long mpow(long long a, long long n, long long mod = MOD){long long res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void pt(){cout << endl; return;}
template<class Head> void pt(Head&& head){cout << head; pt(); return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
void dpt(){if(!isDebugMode) return; cout << endl; return;}
template<class Head> void dpt(Head&& head){if(!isDebugMode) return; cout << head; pt(); return;}
template<class Head, class... Tail> void dpt(Head&& head, Tail&&... tail){if(!isDebugMode) return; cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void enu(T v){rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void enu2(T v){rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T> void debug(T v){if(!isDebugMode) return; rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){if(!isDebugMode) return; rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}
template<class T1, class T2> long long recgcd(T1 a, T2 b){return a % b ? recgcd(b , a % b) : b;}
bool valid(long long H, long long W, long long h, long long w) { return 0 <= h && h < H && 0 <= w && w < W; }

void solve();

int main(){
    while(testcase--) solve();
    return 0;
}

void solve(){
        ll N; cin >> N;
    vector<string> s(N);
    rep(i, N) cin >> s[i];
    ll cnta = 0;
    ll cntb = 0;
    ll cntab = 0;
    ll cntba = 0;
    rep(i, N){
        rep(j, s[i].size() - 1){
            if(s[i][j] == 'A' && s[i][j + 1] == 'B') cntab++;
        }
        if(s[i][0] == 'B' && s[i][s[i].size() - 1] == 'A') cntba++;
        if(s[i][0] == 'B' && s[i][s[i].size() - 1] != 'A') cntb++;
        if(s[i][0] != 'B' && s[i][s[i].size() - 1] == 'A') cnta++;

    }
    // pt(cntab, cnta, cntb);
    ll ans = 0;
    pt(cntab + (cntba == 0 ? min(cnta, cntb) : (cnta + cntb == 0 ? cntba - 1 : cntba + min(cnta, cntb))));
    return;
}