/*

     贪心+构造：因为是对称的，可以全都左一半考虑，过程很简单，但是能想到就很难了

 */

 /************************************************

 Author        :Running_Time

 Created Time  :2015-8-3 9:14:02

 File Name     :B.cpp

  *************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN = 1e5 + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 char s[MAXN];

 int a[MAXN];

 int n, p;

 int main(void)    {       //Codeforces Round #277 (Div. 2) C. Palindrome Transformation

     scanf ("%d%d", &n, &p); scanf ("%s", s + );

     if (p > n / )  p = n - p + ;

     int l = n + , r = ;   int ans = ;

     for (int i=; i<=n/; ++i)    {

         if (s[i] != s[n-i+]) {

             l = min (l, i); r = max (r, i);

             ans += min (abs (s[i] - s[n-i+]),  - abs (s[i] - s[n-i+]));

         }

     }

     if (ans == )   puts ("");

     else    {

         printf ("%d\n", ans + r - l + min (abs (p - l), abs (r - p)));

     }

     return ;

 }