Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and s doesn't contain any palindrome contiguous substring of length 2 or more.

Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.

Input

The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting of n small English letters. It is guaranteed that the string is tolerable (according to the above definition).

Output

If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).

Examples

Input

3 3
cba


Output

NO


Input

3 4
cba


Output

cbd


Input

4 4
abcd


Output

abda

Note

String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.

The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.

A palindrome is a string that reads the same forward or reversed.
import java.util.Stack;
import java.util.HashMap;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.StringTokenizer;
import java.util.ArrayList;

public class C2
{
    static char[]    s;

    static char[]    prefix;

    static boolean[] initial;

    static int[]     initVal;

    static String lastSave = "";


    public static void main(String[] args)
    {
        Scanner in = new Scanner();

        int n = in.nextInt();
        int p = in.nextInt();
        s = in.readNextLine().toCharArray();

        prefix = new char[n];
        initial = new boolean[n];

        for (int i = 0; i < n; i++)
        {
            prefix[i] = s[i];
            initial[i] = true;
        }

        int[] maxes = new int[s.length]; // start point for words
        for (int i = 0; i < maxes.length; i++)
        {
            maxes[i] = s[i] - 'a';
        }

        if (n == 1)
        {
            int ind = s[0] - 'a';
            if (ind < p - 1 && ind != -1)
            {
                System.out.println((char)(ind + 'a' + 1));
            }
            else
            {
                System.out.println("NO");
            }

            return;
        }

        // generate new string lexographically higher up
        // make sure its not a palindrome
        genS(p, maxes, n, n - 1, false);

        System.out.println("NO");
    }


    // level is elements in suffix array
    public static void genS(int p, int[] maxes, int length, int level, boolean increase)
    {
        // full length word is formed
        if (level == length)
        {
            String temp = new String(prefix);
            if (temp.compareTo(new String(s)) > 0)
            {
                if (!isPalindrome(temp))
                {
                    System.out.println(temp);
                    System.exit(0);
                }
            }

            return;
        }

        if (isPalindrome(new String(prefix).substring(0, level)))
        {
            genS(p, maxes, length, level - 1, false);
        }
        else
        // continue word generation
        {
            if (initial[level])
            {
                initial[level] = false;
                for (int i = maxes[level] + 1; i < p; i++)
                {
                    prefix[level] = (char)('a' + i);
                    if (!isPalindrome(new String(prefix).substring(0, level)))
                    {
                        // only continues gen if prefix isn't a palindrome
                        genS(p, maxes, length, level + 1, true);
                    }
                }
            }
            else
            {
                int start = 0;
                if (!increase)
                {
                    start = prefix[level] - 'a';
                }

                for (int i = start; i < p; i++)
                {
                    prefix[level] = (char)('a' + i);
                    if (!isPalindrome(new String(prefix).substring(0, level+1)))
                    {
                        // only continues gen if prefix isn't a palindrome
                        genS(p, maxes, length, level + 1, true);
                    }
                }
            }

            // gets here if none of the letters aren't palindromes
            // breaks off a letter and continues
            if (level == 0)
            {
                // got through all first letter combinations
                System.out.println("NO");
                System.exit(0);
            }
            else
            {
                if (new String(prefix).equals(lastSave))
                {
                    // starting to loop
                    System.out.println("NO");
                    System.exit(0);
                }
                lastSave = new String(prefix).substring(0, level);
                genS(p, maxes, length, level - 1, false);
            }
        }
    }


    // takes in str of any length
    public static boolean isPalindrome(String s)
    {
        if (s.length() < 2)
        {
            return false;
        }

        String twoBefore = String.valueOf(s.charAt(0));
        String oneBefore = String.valueOf(s.charAt(1));

        if (twoBefore.equals(oneBefore))
        {
            return true;
        }

        if (s.length() == 2)
        {
            return false;
        }

        for (int i = 2; i < s.length(); i++)
        {
            String temp = String.valueOf(s.charAt(i));

            if (temp.equals(oneBefore) || temp.equals(twoBefore))
            {
                return true;
            }

            twoBefore = oneBefore;
            oneBefore = temp;
        }

        return false;
    }


    // -----------Pair class for comparisons--------------
    public static class Pair
        implements Comparable<Pair>
    {
        int x, y;


        public Pair(int x, int y)
        {
            this.x = x;
            this.y = y;
        }


        @Override
        public int compareTo(Pair o)
        {
            return Integer.compare(this.x, o.x);
        }


        @Override
        public int hashCode()
        {
            return x ^ y;
        }


        @Override
        public boolean equals(Object o)
        {
            Pair p = (Pair)o;
            return p.x == x && p.y == y;
        }
    }


    // -----------Scanner class for faster input----------
    /*
     * Provides similar API as java.util.Scanner but does not use regular
     * expression engine.
     */
    public static class Scanner
    {
        BufferedReader  br;
        StringTokenizer st;


        public Scanner(Reader in)
        {
            br = new BufferedReader(in);
        }


        public Scanner()
        {
            this(new InputStreamReader(System.in));
        }


        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }


        int nextInt()
        {
            return Integer.parseInt(next());
        }


        long nextLong()
        {
            return Long.parseLong(next());
        }


        double nextDouble()
        {
            return Double.parseDouble(next());
        }


        // Slightly different from java.util.Scanner.nextLine(),
        // which returns any remaining characters in current line,
        // if any.
        String readNextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    // --------------------------------------------------------
}
