Codeforces Round #265 (Div. 2)-C. No to Palindromes!

原题链接

C. No to Palindromes!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 3cba

output

NO

input

3 4cba

output

cbd

input

4 4abcd

output

abda

对于每个字符若str[i] != str[i-1] && str[i] != str[i-2]则必定不会出存在回文串.从尾到头遍历字符串str，对于str[i], 若存在字符变量p使得p > str[i] && p != str[i-1] && p != str[i-2]则str[i] = p, 从j = i + 1开始找到最小的字符，为str[j]赋值满足不为回文串的条件

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <map>#include <vector>#include <queue>#define maxn 1005#define MOD 1000000007#define INF 1e9using namespace std;typedef long long ll;char str[maxn];int main(){//freopen("in.txt", "r", stdin);int n, p;scanf("%d%d", &n, &p);scanf("%s", str);int sign = -1;for(int i = n-1; i >= 0; i--){   for(int j = str[i] + 1; j < p + 'a'; j++){    if(i && j == str[i-1])      continue;    if(i > 1 && j == str[i-2])      continue;    str[i] = j;    sign = i;    break;       }       if(sign != -1)        break;}if(sign == -1){puts("NO");return 0;}for(int i = sign+1; i < n; i++){for(int j = 'a'; j < p + 'a'; j++){if(i && j == str[i-1]) continue;if(i > 1 && j == str[i-2]) continue;str[i] = j;break;}}puts(str);return 0;}