You are given a string q. A sequence of k strings s1, s2, ..., sk is called beautiful, if the concatenation of these strings is string q (formally, s1 + s2 + ... + sk = q) and the first characters of these strings are distinct.

Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

Input

The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence. 

The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

Output

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next k lines print the beautiful sequence of strings s1, s2, ..., sk.

If there are multiple possible answers, print any of them.

Examples

Input

1
abca


Output

YES
abca


Input

2
aaacas


Output

YES
aaa
cas


Input

4
abc


Output

NO

Note

In the second sample there are two possible answers: {"aaaca", "s"} and {"aaa", "cas"}.
#include <bits/stdc++.h>
char str[1100], mstr[1100];
int a[1100];
int main() {
  int n, i, len, j, k, f;
  while (scanf("%d", &n) != EOF) {
    scanf("%s", str);
    len = strlen(str);
    if (n > len)
      printf("NO\n");
    else {
      j = 0;
      for (i = 0; str[i] != '\0'; i++) {
        f = 0;
        if (str[i] != str[i - 1]) {
          for (k = 0; k < j; k++)
            if (str[i] == mstr[k]) {
              f = 1;
              break;
            }
          if (f == 0) {
            mstr[j] = str[i];
            a[j] = i;
            j++;
          }
        }
      }
      if (j < n)
        printf("NO\n");
      else {
        printf("YES\n");
        k = 1;
        for (i = 0; str[i] != '\0'; i++) {
          if (i == a[k] && k < n) {
            printf("\n");
            k++;
          }
          printf("%c", str[i]);
        }
        printf("\n");
      }
    }
  }
  return 0;
}
