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.
public class A {

	public A () {
		int N = sc.nextInt();
		int P = sc.nextInt();
		char [] C = sc.nextChars();

		if (!inc (C, N-1, P))
			exit("NO");
		while (test(C, P) == 0);
		if (test(C, P) == 1)
			print(new String(C));
		else
			print("NO");
	}

	int test(char [] C, int P) {
		int N = C.length;
		for (int i : rep(2, N))
			if (C[i] == C[i-1] || C[i] == C[i-2]) {
				boolean ok = inc(C, i, P);
				if (ok) return 0;
				else return 2;
			}
		return 1;
	}

	boolean inc(char [] C, int i, int P) {
		int N = C.length;
		char Z = (char)('a' + P - 1);
		do
			if (C[i] < Z)
				++C[i];
			else if (i == 0)
				return false;
			else
				return inc(C, i-1, P);
		while ((i >= 1 && C[i] == C[i-1]) || (i >= 2 && C[i] == C[i-2]));

		if (i+1 < N) {
			char x, y, z;
			for (x = 'a'; x <= 'c'; ++x)
				if (x != C[i] && (i == 0 || x != C[i-1]))
					break;
			for (y = 'a'; y <= 'c'; ++y)
				if (y != x && y != C[i])
					break;
			for (z = 'a'; z <= 'c'; ++z)
				if (z != x && z != y)
					break;
			char [] Q = { x, y, z }; int q = 0;
			for (int j : rep(i+1, N))
				C[j] = Q[(q++)%3];
		}

		return true;
	}

	private static int [] rep(int S, int T) { if (T <= S) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	////////////////////////////////////////////////////////////////////////////////////
	private final static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static void print (Object o, Object ... A) { IOUtils.print(o, A); }
	private static void exit (Object o, Object ... A) { IOUtils.print(o, A); IOUtils.exit(); }
	private static class IOUtils {
		public static class MyScanner {
			public String next() { newLine(); return line[index++]; }
			public int nextInt() { return Integer.parseInt(next()); }
			public char [] nextChars() { return next ().toCharArray (); }
			//////////////////////////////////////////////
			private boolean eol() { return index == line.length; }
			private String readLine() {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine() {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split(String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build(Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim(String delim, Object o, Object ... A) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : A)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static void start() { if (t == 0) t = millis(); }
		private static void append(StringBuilder b, Object o, String delim) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					append(b, java.lang.reflect.Array.get(o, i), delim);
			} else if (o instanceof Iterable<?>)
				for (Object p : (Iterable<?>) o)
					append(b, p, delim);
			else {
				if (o instanceof Double)
					o = new java.text.DecimalFormat("#.############").format(o);
				b.append(delim).append(o);
			}
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void print(Object o, Object ... A) { pw.println(build(o, A)); }
		private static void err(Object o, Object ... A) { System.err.println(build(o, A)); }
		private static void exit() {
			IOUtils.pw.close();
			System.out.flush();
			err("------------------");
			err(IOUtils.time());
			System.exit(0);
		}
		private static long t;
		private static long millis() { return System.currentTimeMillis(); }
		private static String time() { return "Time: " + (millis() - t) / 1000.0; }
	}
	public static void main (String[] args) { new A(); IOUtils.exit(); }
}
