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>
using namespace std;
int main() {
  int k;
  scanf("%d", &k);
  int L1;
  char s1[105];
  int cont1[26];
  cin >> s1;
  L1 = strlen(s1);
  memset(cont1, 0, sizeof cont1);
  for (int i = 0; i < L1; ++i) ++cont1[s1[i] - 'a'];
  int c = 0;
  for (int i = 0; i < 26; i++)
    if (cont1[i] != 0) c++;
  for (int i = 0; i < 26; i++)
    if (cont1[i] != 0) cont1[i] = 1;
  if (c < k)
    printf("\nNO");
  else {
    printf("\nYES");
    for (int i = 0; i < L1; ++i) {
      if (cont1[s1[i] - 'a'] == 1 && k > 0) {
        cout << endl << s1[i];
        cont1[s1[i] - 'a'] = 0;
        k--;
      } else
        cout << s1[i];
    }
  }
  return 0;
}
