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;
  vector<string> strings;
  int visited[26] = {0};
  int ans = 1;
  string new_string = "";
  new_string += s[0];
  visited[s[0] - 'a']++;
  int flag = 0;
  if (k == 1) {
    cout << "YES" << endl;
    cout << s;
    flag = 1;
    return 0;
  }
  for (int i = 1; i < s.size(); i++) {
    if (ans >= k) {
      string last_string = new_string;
      for (int j = i; j < s.size(); j++) {
        last_string += s[j];
      }
      strings.push_back(last_string);
      flag = 1;
      cout << "YES" << endl;
      for (int z = 0; z < strings.size(); z++) {
        cout << strings[z] << endl;
      }
      break;
    }
    if (!visited[s[i] - 'a']) {
      strings.push_back(new_string);
      new_string = "";
      new_string += s[i];
      ans++;
      visited[s[i] - 'a']++;
    } else {
      new_string += s[i];
    }
  }
  if (flag == 0) {
    if (ans >= k) {
      string last_string = new_string;
      strings.push_back(last_string);
      flag = 1;
      cout << "YES" << endl;
      for (int z = 0; z < strings.size(); z++) {
        cout << strings[z] << endl;
      }
    } else
      cout << "NO";
  }
  return 0;
}
