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 q;
  scanf("%d", &q);
  char a[100001];
  scanf("%s", a);
  int l = strlen(a);
  if (q > l)
    printf("NO\n");
  else {
    int i = 0;
    int k = 0;
    int b[1000];
    int j = 0;
    b[j++] = i;
    i++;
    while (1) {
      if (i == l) break;
      int c = 1;
      for (k = 0; k < j; k++) {
        if (a[i] == a[b[k]]) {
          c = 0;
          break;
        } else {
        }
      }
      if (c == 1) {
        b[j++] = i;
        i++;
      } else
        i++;
    }
    if (j >= q) {
      printf("YES\n");
      for (i = 0; i < q - 1; i++) {
        for (k = b[i]; k < b[i + 1]; k++) {
          printf("%c", a[k]);
        }
        printf("\n");
      }
      k = b[i];
      for (i = k; i < l; i++) printf("%c", a[i]);
    } else
      printf("NO\n");
  }
  return 0;
}
