One-Dimensional Maze

BaoBao is trapped in a one-dimensional maze consisting of grids arranged in a row! The grids are numbered from 1 to from left to right, and the -th grid is marked with a character , where is either 'L' or 'R'.

Starting from the -th grid, BaoBao will repeatedly take the following steps until he escapes the maze:

• If BaoBao is in the 1st grid or the -th grid, then BaoBao is considered to arrive at the exit and thus can escape successfully.
• Otherwise, let BaoBao be in the -th grid. If , BaoBao will move to the -th grid; If , Baobao will move to the -th grid.

Before taking the above steps, BaoBao can change the characters in some grids to help himself escape. Concretely speaking, for the -th grid, BaoBao can change from 'L' to 'R', or from 'R' to 'L'.

But changing characters in grids is a tiring job. Your task is to help BaoBao calculate the minimum number of grids he has to change to escape the maze.

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 two integers and (, ), indicating the number of grids in the maze, and the index of the starting grid.

The second line contains a string () consisting of characters 'L' and 'R'. The -th character of indicates the character in the -th grid.

It is 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 minimum number of grids BaoBao has to change to escape the maze.

Sample Input

3
3 2
LRL
10 4
RRRRRRRLLR
7 4
RLLRLLR

Sample Output

0
2
1

Hint

For the first sample test case, BaoBao doesn't have to change any character and can escape from the 3rd grid. So the answer is 0.

For the second sample test case, BaoBao can change to 'R' and to 'R' and escape from the 10th grid. So the answer is 2.

For the third sample test case, BaoBao can change to 'L' and escape from the 1st grid. So the answer is 1.

題目連結：

https://vjudge.net/problem/ZOJ-3992

題意描述：

給你一個長度為n的字串迷宮，然後給你一個被困的位置，該位置為第幾個字母，如果向右走則L為牆，如果向左走R為牆，問向那個方向走穿過的牆少，到達迷宮的邊緣時則認為走出迷宮邊緣即為字串的首字元和尾字元

解題思路：

把這些字元存在一個字串中，然後從第m個字元開始向兩個方向走，最後輸出穿過牆數較少的方向的牆的數目

程式程式碼：

#include<stdio.h>
#include<algorithm>
using namespace std;
char str[100010];

int main()
{
	int i,j,T,n,m,sum1,sum2;
	
	scanf("%d",&T);
	while(T--)
	{
		sum1=0;
		sum2=0;
		scanf("%d%d",&n,&m);
		scanf("%s",str);
		for(i=m-1;i<n-1;i++)
			if(str[i]!='R')
				sum1++;
		for(i=m-1;i>0;i--)
			if(str[i]!='L')
				sum2++;
		printf("%d\n",min(sum1,sum2));
	}
    return 0;
}