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() {
  char a[105];
  int x, b[30] = {0}, len = 0;
  scanf("%d%s", &x, a);
  len = strlen(a);
  int num = 0;
  for (int i = 0; i < len; i++) {
    if (b[a[i] - 'a'] == 0) {
      num++;
      b[a[i] - 'a']++;
    }
  }
  if (x > len || num < x) {
    printf("NO\n");
  } else if (x == 1) {
    printf("YES\n%s\n", a);
  } else {
    printf("YES\n");
    int t = 0;
    int b[30] = {0};
    for (int i = 0; i < x - 1; i++) {
      char k = a[t];
      for (int j = t; j < len; j++) {
        if (a[j] != k && b[a[j] - 'a'] == 0) {
          printf("\n");
          t = j;
          break;
        } else {
          printf("%c", a[j]);
          if (j == t) b[k - 'a']++;
        }
      }
    }
    for (int i = t; i < len; i++) {
      printf("%c", a[i]);
    }
    printf("\n");
  }
  return 0;
}
