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 x, point[100];
char a[100];
void print_point(int l) {
  int i;
  for (i = 0; i <= l; i++) printf("%d\t", point[i]);
  printf("\n");
}
void print(int start, int end) {
  if (start > end) return;
  if (end > x - 1) end = x;
  int i = start;
  for (i = start; i <= end; i++) printf("%c", a[i]);
  printf("\n");
  return;
}
int main() {
  int i, k, count = 1, alpha[26], mm = 1;
  char temp;
  scanf("%d", &k);
  scanf("%s", a);
  memset(alpha, 0, sizeof(alpha));
  point[0] = 0;
  x = strlen(a);
  i = 1;
  alpha[a[0] - 'a'] = 1;
  temp = a[0];
  if (k == 1) {
    printf("YES\n%s\n", a);
  } else {
    while (i < x) {
      if (count >= k) {
        break;
      }
      if (a[i] != temp && alpha[a[i] - 'a'] == 0) {
        count++;
        point[mm++] = i;
        temp = a[i];
        alpha[a[i] - 'a'] = 1;
      }
      i++;
    }
    mm--;
    if (count < k)
      printf("NO\n");
    else {
      i = 1;
      printf("YES\n");
      print(0, point[1] - 1);
      while (i < mm) {
        print(point[i], point[i + 1] - 1);
        i++;
      }
      print(point[i], x - 1);
    }
  }
  return 0;
}
