No to Palindromes

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 s ₁ = t ₁, …, 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.

Solution:

#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

char s[1234567];

int main() {
  int n, p;
  scanf("%d %d", & n, & p);
  scanf("%s", s);
  for (int i = n - 1; i >= 0; i--) {
    for (char c = s[i] + 1; c < 'a' + p; c++) {
      if (i - 1 >= 0 && s[i - 1] == c) {
        continue;
      }
      if (i - 2 >= 0 && s[i - 2] == c) {
        continue;
      }
      if (p == 2 && n > 2) {
        continue;
      }
      if (p == 1 && n > 1) {
        continue;
      }
      for (int j = 0; j < n; j++) {
        if (j < i) {
          putchar(s[j]);
          continue;
        }
        if (j == i) {
          s[i] = c;
          putchar(s[i]);
          continue;
        }
        for (char c = 'a'; c < 'a' + p; c++) {
          if (j - 1 >= 0 && s[j - 1] == c) {
            continue;
          }
          if (j - 2 >= 0 && s[j - 2] == c) {
            continue;
          }
          s[j] = c;
          break;
        }
        putchar(s[j]);
      }
      printf("\n");
      return 0;
    }
  }
  puts("NO");
  return 0;
}