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>
int main() {
  int n;
  char s[102];
  int a[30] = {0};
  scanf("%d%s", &n, &s);
  int len = strlen(s);
  int i;
  int sum = 0;
  for (i = 0; i < len; i++) {
    if (a[s[i] - 'a' + 1] == 0) sum++, a[s[i] - 'a' + 1] = 1;
  }
  if (sum >= n)
    printf("YES\n");
  else {
    printf("NO\n");
    return 0;
  }
  printf("%c", s[0]);
  a[s[0] - 'a' + 1] = 0;
  sum = 1;
  for (i = 1; i < len; i++) {
    if (sum == n) {
      printf("%c", s[i]);
      continue;
    }
    if (a[s[i] - 'a' + 1] == 0) {
      printf("%c", s[i]);
      continue;
    }
    if (sum < n && a[s[i] - 'a' + 1] == 1) {
      sum++;
      printf("\n%c", s[i]);
      a[s[i] - 'a' + 1] = 0;
      continue;
    }
  }
  printf("\n");
}
