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 k, i, j, s[28], n[28];
  char str[105];
  scanf("%d", &k);
  scanf("%s", str);
  for (i = 1; i <= 26; i++) {
    s[i] = 1;
    n[i] = 0;
  }
  int l = strlen(str);
  s[str[0] - 96] = 0, n[1] = 0;
  j = 1;
  for (i = 1; i < l; i++) {
    if (s[str[i] - 96] == 1) {
      s[str[i] - 96] = 0;
      j++;
      n[j] = i;
    }
  }
  if (j < k)
    printf("NO\n");
  else {
    printf("YES\n");
    for (i = 1; i < k; i++) {
      for (int u = n[i]; u <= n[i + 1] - 1; u++) printf("%c", str[u]);
      printf("\n");
    }
    for (int u = n[k]; u <= l - 1; u++) printf("%c", str[u]);
    printf("\n");
  }
  return 0;
}
