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 a[26] = {0}, len, n, t = 0, b[100], i, j;
  char s[105];
  scanf("%d", &n);
  scanf("%s", s);
  len = strlen(s);
  if (len < n) {
    printf("NO\n");
    return 0;
  }
  if (n == 1) {
    printf("YES\n");
    printf("%s", s);
    return 0;
  }
  for (i = 0; s[i] != '\0'; i++) {
    if (!a[s[i] - 'a']) {
      a[s[i] - 'a'] = 1;
      b[t++] = i;
      if (t == n) {
        b[t] = len - 1;
        break;
      }
    }
  }
  if (b[t] != len - 1) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (i = 0; i < n - 1; i++) {
    for (j = b[i]; j < b[i + 1]; j++) printf("%c", s[j]);
    printf("\n");
  }
  for (i = b[n - 1]; i <= b[n]; i++) printf("%c", s[i]);
}
