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;
  cin >> n;
  string s;
  cin >> s;
  map<char, int> mp;
  if (s.size() < n) {
    cout << "NO" << endl;
  } else {
    int counter = 1;
    string str = "";
    int flag = 0;
    vector<string> vi;
    mp[s[0]] = 1;
    str += s[0];
    if (n == 1) {
      cout << "YES" << endl;
      cout << "" << s << endl;
    } else {
      for (int i = 1; i < s.size(); i++) {
        if (mp[s[i]] == 0) {
          mp[s[i]] = 1;
          vi.push_back(str);
          str = "";
          str += s[i];
          counter++;
        } else {
          str += s[i];
        }
        if (counter == n) {
          flag = 1;
          str = str + s.substr(i + 1);
          vi.push_back(str);
          break;
        }
      }
      if (flag == 1) {
        cout << "YES" << endl;
        for (int i = 0; i < vi.size(); i++) {
          cout << "" << vi[i] << endl;
        }
      } else {
        cout << "NO" << endl;
      }
    }
  }
}
