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 n, k = 0;
  char c[101];
  scanf("%d%s", &n, &c[0]);
  bool f[30];
  for (int i = 0; i < 30; i++) {
    f[i] = true;
  }
  for (int i = 0; c[i] >= 'a' && c[i] <= 'z'; i++) {
    if (f[c[i] - 'a']) {
      k++;
      f[c[i] - 'a'] = false;
    }
  }
  for (int i = 0; i < 30; i++) {
    f[i] = true;
  }
  if (k >= n) {
    printf("YES\n");
    f[c[0] - 'a'] = false;
    for (int i = 0; c[i] >= 'a' && c[i] <= 'z'; i++) {
      if (f[c[i] - 'a'] && n > 1) {
        n--;
        printf("\n");
        f[c[i] - 'a'] = false;
      }
      printf("%c", c[i]);
    }
  } else {
    printf("NO");
  }
  return 0;
}
