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 ar[30];
int main() {
  int n;
  scanf("%d\n", &n);
  string s;
  getline(cin, s);
  vector<string> res;
  string t;
  int n2 = n;
  t.push_back(s[0]);
  ar[s[0] - 'a'] = 1;
  n2--;
  for (int i = 1; i < s.size(); i++) {
    if (n2 == 0)
      t.push_back(s[i]);
    else if (ar[s[i] - 'a'] == 0) {
      res.push_back(t);
      t.clear();
      t.push_back(s[i]);
      ar[s[i] - 'a'] = 1;
      n2--;
    } else
      t.push_back(s[i]);
  }
  if (n2 != 0)
    printf("NO\n");
  else {
    res.push_back(t);
    printf("YES\n");
    for (int i = 0; i < res.size(); i++) printf("%s\n", res[i].c_str());
  }
  return 0;
}
