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 get_string(string s, int from, int to) {
  string p;
  for (int i = from; i <= to; i++) p += s[i];
  return p;
}
int k;
string s, d;
vector<int> v;
bool used[32];
int i;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> k;
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (!used[s[i] - 'a']) {
      used[s[i] - 'a'] = true;
      v.push_back(i);
    }
  }
  if (v.size() < k) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  for (i = 0; i < k - 1; i++) {
    cout << get_string(s, v[i], v[i + 1] - 1) << '\n';
  }
  for (i = k - 1; i < v.size(); i++) {
    if (i == v.size() - 1)
      d += get_string(s, v[i], s.size() - 1);
    else
      d += get_string(s, v[i], v[i + 1] - 1);
  }
  cout << d << '\n';
  return 0;
}
