Paul hates palindromes. He assumes that strings is tolerable if each its character is one of the firstp 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 lengthn. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.

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

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

String s is lexicographically larger (or simply larger) than stringt with the same length, if there is numberi, such that s₁?=?t₁, ...,s_i?=?t_i,s_i?+?1?>?t_i?+?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.

题意：求满足大于给定串的序列，而且两个序列都不包含长度大于等于2的回文子串

思路：首先，因为不能有长度大于等于2的子串，所以当前位置是不能和它的前一个和前第二个一样的。因为给定串也是不包含回文子串的，所以为了能找个一个最小的满足条件的字符串，我们从右往前改，找到一个满足不与前两个一样的位置的话，那么从这个位置往右就是填写不构成回文的最小串了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;

int n, p;
char str[maxn], tmp;

int find(const int &x) {
	if (x == -1) 
		return 0;
	for (int i = 1; ; i++) {
		if (str[x] + i >= tmp) break;
		if (x > 0 && str[x-1] == str[x] + i) continue;
		if (x > 0 && str[x-2] == str[x] + i) continue;
		str[x] += i;
		return 1;
	}

	if (!find(x-1)) 
		return 0;

	str[x] = 'a';
	for (int i = 0; ; i++) {
		if (str[x] + i >= tmp) break;
		if (x > 0 && str[x-1] == str[x] + i) continue;
		if (x > 1 && str[x-2] == str[x] + i) continue;
		str[x] += i;
		return 1;
	}
	return 0;
}

int main() {
	scanf("%d%d", &n, &p);	
	tmp = 'a' + p;
	scanf("%s", str);
	if (find(n-1))
		printf("%s\n", str);
	else printf("NO\n");
	return 0;
}