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() {
  string s;
  int k;
  bool exist[26];
  vector<int> pos;
  while (cin >> k) {
    cin >> s;
    if (k == 1) {
      cout << "YES" << endl;
      cout << s << endl;
      continue;
    }
    memset(exist, 0, sizeof(exist));
    exist[(int)(s[0] - 'a')] = true;
    pos.clear();
    pos.push_back(0);
    int n = s.length();
    for (int i = 1; i < n; i++) {
      int j = (int)(s[i] - 'a');
      if (exist[j]) continue;
      exist[j] = true;
      pos.push_back(i);
      if (pos.size() == k) break;
    }
    if (pos.size() < k) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      for (int i = 0; i < pos.size(); i++) {
        int start = pos[i];
        int end = (i < pos.size() - 1) ? pos[i + 1] : n;
        cout << s.substr(start, end - start) << endl;
      }
    }
  }
  return 0;
}
