ZOJ

One-Dimensional Maze

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Time Limit: 1 Second      Memory Limit: 65536 KB

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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

33 2LRL10 4RRRRRRRLLR7 4RLLRLLR

Sample Output

021

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.

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int N = 1e5 + 10;int T,n,m;char s[N];int main(){    scanf("%d",&T);    while(T--){        scanf("%d%d%s",&n,&m,s+1);        int len = strlen(s+1), ans1 = 0, ans2 = 0;        for(int i=2;i<=m;i++){            if(s[i]=='R') ans1++;        }        for(int i=m;i<n;i++){            if(s[i]=='L') ans2++;        }        printf("%d\n",min(ans1, ans2));    }    return 0;}