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;
  string s;
  scanf("%d", &n);
  cin >> s;
  int used[27];
  vector<int> from;
  memset(used, 0, sizeof(used));
  for (int i = 0; i < s.length(); i++) {
    if (used[s[i] - 'a'] == 0) {
      used[s[i] - 'a'] = 1;
      from.push_back(i);
    }
  }
  from.push_back(s.length());
  int cnt = from.size();
  if (cnt >= n + 1) {
    printf("YES\n");
    int curr = 0;
    int pos = 0;
    while (curr < n) {
      while (pos < from[curr + 1]) {
        printf("%c", s[pos]);
        pos++;
      }
      if (curr != n - 1) printf("\n");
      curr++;
    }
    while (pos < s.length()) {
      printf("%c", s[pos]);
      pos++;
      if (pos == s.length()) printf("\n");
    }
  } else {
    printf("NO\n");
  }
  return 0;
}
