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 aa[30], bb[30];
char s[105];
int main() {
  int m = 0, k, n;
  scanf("%d%s", &k, s);
  n = strlen(s);
  for (int i = 0; i < n; i++)
    if (aa[s[i] - 'a'] == 0) {
      m++;
      aa[s[i] - 'a'] = 1;
    }
  if (m < k || n < k)
    printf("NO");
  else {
    memset(aa, 0, sizeof(aa));
    aa[s[0] - 'a'] = 1;
    m = 1;
    for (int i = 1; i < n; i++) {
      if (m == k) break;
      if (aa[s[i] - 'a'] == 0) {
        aa[s[i] - 'a'] = 1;
        bb[m] = i;
        m++;
      }
    }
    bb[m++] = n;
    printf("YES\n");
    for (int i = 0; i < k; i++) {
      for (int j = bb[i]; j < bb[i + 1]; j++) printf("%c", s[j]);
      printf("\n");
    }
  }
  return 0;
}
