G. ABBC or BACB

You are given a string $s$ made up of characters $\texttt{A}$ and $\texttt{B}$. Initially you have no coins. You can perform two types of operations:

Pick a substring$^\dagger$ $\texttt{AB}$, change it to $\texttt{BC}$, and get a coin.
Pick a substring$^\dagger$ $\texttt{BA}$, change it to $\texttt{CB}$, and get a coin.

What is the most number of coins you can obtain?
$^\dagger$ A substring of length $2$ is a sequence of two adjacent characters of a string.

Input

The input consists of multiple test cases. The first line of the input contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The only line of each test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). All characters of $s$ are either $\texttt{A}$ or $\texttt{B}$.

The sum of the lengths of $s$ over all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output a single integer — the maximum number of coins you can obtain.

Example

input

8
ABBA
ABA
BAABA
ABB
AAAAAAB
BABA
B
AAA

output

Note

In the first test case you can perform the following operations to get $2$ coins: $${\color{red}{\texttt{AB}}}\texttt{BA} \to \texttt{BC}{\color{red}{\texttt{BA}}} \to \texttt{BCCB}$$

In the second test case you can perform the following operation to get $1$ coin: $${\color{red}{\texttt{AB}}}\texttt{A} \to \texttt{BCA}$$

In the third test case you can perform the following operations to get $3$ coins: $${\color{red}{\texttt{BA}}}\texttt{ABA} \to \texttt{CBA}{\color{red}{\texttt{BA}}} \to \texttt{C}{\color{red}{\texttt{BA}}}\texttt{CB} \to \texttt{CCBCB}$$

解题思路

经典思维题不会做.jpg

官方给出的题解很长，这里简练下给出其核心思想。最关键的地方是要注意到任意一种操作执行后字符 $\texttt{B}$ 的变化情况。

如果是 $\texttt{AB} \to \texttt{BC}$，那么可以理解成 $\texttt{B}$ 左移并覆盖其左邻的 $\texttt{A}$，再用 $\texttt{C}$ 覆盖原来的位置。同理对于 $\texttt{BA} \to \texttt{CB}$，可以理解成 $\texttt{B}$ 右移并覆盖其右邻的 $\texttt{A}$，再用 $\texttt{C}$ 覆盖原来的位置。

因此对于 $\texttt{AAAB}$ 这一类字符串，将 $\texttt{B}$ 不断左移并覆盖，即有 $\texttt{AAAB} \to \texttt{AABC} \to \texttt{ABCC} \to \texttt{BCCC}$，那么就可以获得最大收益（即操作次数最多）。同理对于 $\texttt{BAAA}$ 这一类字符串，将 $\texttt{B}$ 不断右移并覆盖，即有 $\texttt{BAAA} \to \texttt{CBAA} \to \texttt{CCBA} \to \texttt{CCCB}$，可以获得最大收益。

因此可以发现对于一个字符 $\texttt{B}$，如果其两边相邻的都是 $\texttt{A}$，那么可以选择左移或右移，并且之后只能向同个方向移动（因为另一个方向会变成 $\texttt{C}$）。因此一个字符 $\texttt{B}$ 只能覆盖某一段与其相邻且连续的 $\texttt{A}$。所以我们找到字符串中所有的 $\texttt{B}$，假设有 $k$ 个，那么就会将整个字符串中的 $\texttt{A}$ 分成 $k+1$ 个段（包括空串）。例如有 $\texttt{AABBAAAB}$，那么 $\texttt{A}$ 就被分成 $\left\{ \texttt{AA}, \, \phi, \, \texttt{AAA}, \phi \right\}$。由于 $k$ 个 $\texttt{B}$ 只能覆盖 $k$ 段的 $\texttt{A}$，意味着必然有一段 $\texttt{A}$ 无法被覆盖，而为了得到最大收益，我们选择长度最小的那段舍弃即可。

因此我们只需根据 $\texttt{B}$ 的划分来累加每一段 $\texttt{A}$ 的长度，并维护长度的最小值。最后答案就是 $\texttt{A}$ 的个数减去这个最小值。

AC 代码如下，时间复杂度为 $O(n)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

char s[N];

void solve() {
	scanf("%s", s);
	int sum = 0, mn = 1e9;
	for (int i = 0, j = -1; s[i]; i++) {
		if (s[i] == 'B') {
			int t = i - j - 1;
			sum += t;
			mn = min(mn, t);
			j = i;
		}
		if (!s[i + 1]) {
			int t = i - j;
			sum += t;
			mn = min(mn, t);
		}
	}
	printf("%d\n", sum - mn);
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		solve();
	}
	
	return 0;
}