对于连续的 $A$，可以使用左边的 $B$ 消除，同时把 $B$ 右移；也可以使用右边的 $B$ 消除，同时把 $B$ 左移。

先考虑什么情况下不能完全消除，字符串的形式应该类似这样：$AAABAAABAAABAAA$，也就是 $x$ 个单独的 $B$，把所有的 $A$ 分成了 $x + 1$ 段，此时只能消除 $x$ 段 $A$，为了获取最大价值，应该留下最小的那段连续的 $A$。

然后考虑什么情况下可以完全消除，只需要在上面的字符串的基础上，添加一个「多余的」$B$ 即可，它的作用只是用来消除最后剩下的那段 $A$，添加方式有三种：在字符串的最开始，在字符串的最后，紧挨着现有的 $B$

// Date: Thu Dec 28 22:27:09 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200100;
int n, t, a[N];
string s;

int main(void) {
#ifdef _DEBUG
  freopen("1873g.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> s;
    n = SZ(s);

    int idx = 0, cnt = 0;
    bool flag = false;

    For(i, 0, n) {
      if (s[i] == 'B') {
        if (!flag) {
          if ((i == 0) || (i == n - 1) || (i + 1 < n && s[i + 1] == 'B')) {
            flag = true;
          }
        }
        if (cnt) {
          a[++idx] = cnt;
          cnt = 0;
        }
      } else {
        cnt++;
      }
    }

    if (cnt)
      a[++idx] = cnt;

    ll sum = 0;
    int mi_x = INF;
    For1(i, 1, idx) {
      sum += a[i];
      mi_x = min(mi_x, a[i]);
    }

    if (flag)
      cout << sum;
    else
      cout << (sum - mi_x);
    cout << '\n';
  }

  return 0;
}