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 length(char T[101]) {
  int cont = 0;
  for (; T[cont]; cont++)
    ;
  return cont;
}
int exista_pana_acum(char pana_acum[101], char c) {
  for (int i = 0; pana_acum[i]; i++) {
    if (pana_acum[i] == c) return 1;
  }
  return 0;
}
void afisaza_intre_pana_acum(char T[101], int n1, int n2) {
  for (int i = n1; i < n2 && T[i]; i++) printf("%c", T[i]);
  printf("\n");
}
int main() {
  int k;
  char T[101] = {};
  int pozitii_afisat[101] = {};
  char pana_acum[101] = {};
  scanf("%d", &k);
  scanf(" %[^\n]", T);
  int L = length(T), ultima = 0, siruri_gasite = 0, se_poate = 0;
  if (L < k) {
    printf("NO\n");
    return 0;
  } else {
    for (int i = 0;; i++) {
      if (siruri_gasite == k) {
        se_poate = 1;
        break;
      }
      if (!T[i]) break;
      if (!exista_pana_acum(pana_acum, T[i])) {
        pana_acum[siruri_gasite] = T[i];
        pozitii_afisat[siruri_gasite++] = i;
      }
    }
  }
  if (!se_poate)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < siruri_gasite - 1; i++) {
      afisaza_intre_pana_acum(T, pozitii_afisat[i], pozitii_afisat[i + 1]);
    }
    for (int i = pozitii_afisat[siruri_gasite - 1]; T[i]; i++)
      printf("%c", T[i]);
    printf("\n");
  }
  return 0;
}
