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 k;
string s;
vector<string> ans;
bool all[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> k >> s;
  string build = "";
  for (int i = 0; i < s.size(); ++i)
    if (!all[s[i] - 'a']) {
      if (!build.empty()) ans.push_back(build);
      build = s[i];
      all[s[i] - 'a'] = 1;
    } else
      build += s[i];
  ans.push_back(build);
  if (ans.size() < k)
    cout << "NO";
  else {
    cout << "YES"
         << "\n";
    while (ans.size() > k) {
      int last = ans.size() - 1;
      string s1 = ans[last], s2 = ans[last - 1];
      ans.pop_back();
      ans.pop_back();
      ans.push_back(s2 + s1);
    }
    for (int i = 0; i < ans.size(); ++i) cout << ans[i] << "\n";
  }
  return 0;
}
