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 k;
  cin >> k;
  string s;
  cin >> s;
  int flag[28] = {0};
  int num = 0;
  for (int i = 0; i < s.length(); i++) {
    if (flag[s[i] - 'a'] == 0) {
      flag[s[i] - 'a'] = 1;
      num++;
      if (num == k) {
        break;
      }
    }
  }
  if (num < k) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    int pos = 0;
    int count = 0;
    while (pos < s.length() && count < k) {
      flag[s[pos] - 'a'] = 2;
      int start = pos;
      int send;
      for (send = start + 1; send < s.length(); send++) {
        if (flag[s[send] - 'a'] != 2) {
          break;
        }
      }
      if (send == s.length()) {
        cout << s.substr(start, s.length() - start) << endl;
        break;
      }
      if (flag[s[send] - 'a'] == 0) {
        cout << s.substr(start, s.length() - start) << endl;
      } else {
        if (count < k - 1)
          cout << s.substr(start, send - start) << endl;
        else if (count == k - 1) {
          cout << s.substr(start, s.length() - start) << endl;
        }
      }
      count++;
      pos = send;
    }
  }
  return 0;
}
