String of CCPC

BaoBao has just found a string of length consisting of 'C' and 'P' in his pocket. As a big fan of the China Collegiate Programming Contest, BaoBao thinks a substring of is "good", if and only if 'C', and 'P', where denotes the -th character in string . The value of is the number of different "good" substrings in . Two "good" substrings and are different, if and only if .

To make this string more valuable, BaoBao decides to buy some characters from a character store. Each time he can buy one 'C' or one 'P' from the store, and insert the character into any position in . But everything comes with a cost. If it's the -th time for BaoBao to buy a character, he will have to spend units of value.

The final value BaoBao obtains is the final value of minus the total cost of all the characters bought from the store. Please help BaoBao maximize the final value.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer (), indicating the length of string .

The second line contains the string () consisting of 'C' and 'P'.

It's guaranteed that the sum of over all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the maximum final value BaoBao can obtain.

Sample Input

3
3
CCC
5
CCCCP
4
CPCP

Sample Output

1
1
1

Hint

For the first sample test case, BaoBao can buy one 'P' (cost 0 value) and change to "CCPC". So the final value is 1 - 0 = 1.

For the second sample test case, BaoBao can buy one 'C' and one 'P' (cost 0 + 1 = 1 value) and change to "CCPCCPC". So the final value is 2 - 1 = 1.

For the third sample test case, BaoBao can buy one 'C' (cost 0 value) and change to "CCPCP". So the final value is 1 - 0 = 1.

It's easy to prove that no strategies of buying and inserting characters can achieve a better result for the sample test cases.

題目連結：
https://vjudge.net/problem/ZOJ-3985

題意描述：

給你一個由CP組成的字串，允許你加入一個字母C或P，使字串中的CCPC數量最多，其中CCPCCPC這個串中含有兩個CCPC

解題思路：

首先把所有的CCPC判斷一下，如果不滿足CCPC那就判斷CCC、CCP和CPC，其中CCC還要判斷其後面是否為PC即可

程式程式碼：

#include<stdio.h>
#include<algorithm>
using namespace std;
char str[200010];

int main()
{
	int i,T,n,sum,flag;
	
	scanf("%d",&T);
	while(T--)
	{
		sum=0;
		flag=0;
		scanf("%d%s",&n,str);
		for(i=0;i<n;i++)
		{
			if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='P'&&str[i+3]=='C')
			{
				sum++;
				i+=2;
				continue;
			}
			if(flag==0)
			{
				if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='C')
				{
					if(str[i+3]=='P'&&str[i+4]=='C')
						continue;
					sum++;
					flag=1;
				}
				if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='P')
				{
					sum++;
					flag=1;
				}
				if(str[i]=='C'&&str[i+1]=='P'&&str[i+2]=='C')
				{
					sum++;
					flag=1;
				}
			}
		}
		printf("%d\n",sum);
	}
    return 0;
}