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 k, l, i = 0, j = -1, m = 0;
  scanf("%d", &k);
  char a[101], c[k][101];
  scanf("%s", a);
  int d[k];
  int b[26] = {0};
  l = strlen(a);
  if (l < k)
    printf("NO\n");
  else {
    while (k > 0 && a[i] != '\0') {
      if (b[(int)a[i] - 97] == 0) {
        if (j >= 0) {
          d[j] = m;
          m = 0;
        }
        k--;
        b[(int)a[i] - 97] = 1;
        c[++j][m] = a[i];
        i++;
        if (k == 0) {
          while (a[i] != '\0') {
            c[j][++m] = a[i];
            i++;
          }
          break;
        }
      } else {
        c[j][++m] = a[i];
        i++;
      }
      d[j] = m;
    }
    if (k == 0) {
      printf("YES\n");
      for (i = 0; i <= j; i++) {
        if (i == j) {
          for (l = 0; l <= m; l++) printf("%c", c[i][l]);
          printf("\n");
        } else {
          for (l = 0; l <= d[i]; l++) printf("%c", c[i][l]);
          printf("\n");
        }
      }
    } else
      printf("NO\n");
  }
  return 0;
}
