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

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

• If BaoBao is in the 1st grid or the \(n\)-th grid, then BaoBao is considered to arrive at the exit and thus can escape successfully.
• Otherwise, let BaoBao be in the \(t\)-th grid. If \(s_t = \text{'L'}\), BaoBao will move to the \((t-1)\)-th grid; If \(s_t = \text{'R'}\), Baobao will move to the \((t+1)\)-th grid.

Before taking the above steps, BaoBao can change the characters in some grids to help himself escape. Concretely speaking, for the \(i\)-th grid, BaoBao can change \(s_i\) 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.

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

The first line contains two integers \(n\) and \(m\) (\(3 \le n \le 10^5\), \(1 < m < n\)), indicating the number of grids in the maze, and the index of the starting grid.

The second line contains a string \(s\) (\(|s| = n\)) consisting of characters 'L' and 'R'. The \(i\)-th character of \(s\) indicates the character in the \(i\)-th grid.

It is guaranteed that the sum of \(n\) over all test cases will not exceed \(10^6\).

For each test case output one line containing one integer, indicating the minimum number of grids BaoBao has to change to escape the maze.

3

3 2

LRL

10 4

RRRRRRRLLR

7 4

RLLRLLR

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 \(s_8\) to 'R' and \(s_9\) to 'R' and escape from the 10th grid. So the answer is 2.

For the third sample test case, BaoBao can change \(s_4\) to 'L' and escape from the 1st grid. So the answer is 1.

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

#define max_v 100005

char str[max_v];

/*

Submit Failed

*/

int main()

{

    int t;

    int n,m,sum1,sum2;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d %d",&n,&m);

        scanf("%s",str+);

        sum1=;

        sum2=;

        for(int i=m;i<n;i++)

            if(str[i]=='L')

               sum1++;

        for(int i=m;i>;i--)

            if(str[i]=='R')

            sum2++;

        printf("%d\n",min(sum1,sum2));

    }

    return ;

}

/*

题意：给出一串只有RL的字符串，

给定起始位置，R表示向右，L表示向左，

现在要求输出最少改变几个字符可以移动到字符第一个或者最后一个。

例如：

6 3

LLRRLR 

起始位置是第三个字符R，只用改变第三个R就可以到达第一个字符L

这道题目通过模拟就可以做出来，

就是将到达两边要改变的字符都求出来，输出最小的那个 

*/