codeForces 486C Palindrome Transformation

C. Palindrome Transformation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.

There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).

When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.

Initially, the text cursor is at position p.

Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.

The next line contains n lowercase characters of Nam's string.

Output

Print the minimum number of presses needed to change string into a palindrome.

Examples

input

8 3aeabcaez

output

6

Note

A string is a palindrome if it reads the same forward or reversed.

In the sample test, initial Nam's string is:  (cursor position is shown bold).

In optimal solution, Nam may do 6 following steps:

The result, , is now a palindrome.

题意：给出一个字符串，字符串的长度，指针的位置，和四种操作，上下左右，left 是指针左移，right 指针右移，up字母向字典序增大的方向改变如a可以变为b,down是将将字母向字典序减小的方向改变。问经过多少次改变可以将字符串变为一个回文串。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[100000+10];int x[100000+10],y[100000+10];int main(){int n,m,i,j,l,k,r;while(scanf("%d%d",&n,&m)!=EOF){scanf("%s",s+1);if(n==1){printf("0\n");continue;}k=0;int cnt=0;if(n&1){j=n/2+2;}else{j=n/2+1;}for(i=n/2;i>=0&&j<=n;i--,j++){if(s[i]!=s[j]){int a=s[i]-'a';int b=s[j]-'a';int minn=min(a,b);int maxx=max(a,b);cnt+=min(maxx-minn,minn+26-maxx);//寻找字母之间需要变更的最小次数 x[k]=i;//储存需要变更的位置 y[k++]=j;}}sort(x,x+k);sort(y,y+k);if(m<=n/2)//如果指针在回文串的左边 {if(k){if(k==1){cnt+=abs(x[0]-m);}else{int aa=min(abs(m-x[0]),abs(x[k-1]-m));int flag=0;if(m>=x[0]){flag++;cnt+=m-x[0];}//最左边距离指针的距离 if(m<x[k-1]){flag++;cnt+=x[k-1]-m;}//最右边距离指针的距离 if(flag==2)//指针在需要变更的中间位置 cnt+=aa;}}}else//指针在回文串的右边 {    if(k)    {    if(k==1)    cnt+=abs(y[0]-m);    else    {    int aa=min(abs(m-y[0]),abs(y[k-1]-m));int flag=0;if(m>=y[0]){flag++;cnt+=m-y[0];}if(m<y[k-1]){flag++;cnt+=y[k-1]-m;}if(flag==2)cnt+=aa;}}}printf("%d\n",cnt);}return 0;}