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;
string s;
int k, l, p;
map<char, bool> chk, boobs;
int main() {
  chk = map<char, bool>();
  boobs = map<char, bool>();
  cin >> k;
  cin >> s;
  l = s.length();
  for (int i = 0; i < l; i++) {
    if (!chk[s.at(i)]) {
      chk[s.at(i)] = true;
      p++;
    }
  }
  if (p < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  int i = 0;
  while (k) {
    if (k == 1) {
      while (i < l) {
        cout << s.at(i);
        i++;
      }
      cout << "\n";
      k--;
    } else {
      if (!boobs[s.at(i)] && i) {
        cout << "\n";
        k--;
      }
      boobs[s.at(i)] = true;
      cout << s.at(i);
      i++;
    }
  }
  return 0;
}
