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;
inline long double min(long double a, long double b) {
  if (a < b) return a;
  return b;
}
inline long double max(long double a, long double b) {
  if (a < b) return b;
  return a;
}
int n;
string s;
unordered_set<char> umap;
int main() {
  cin >> n;
  cin >> s;
  for (int(i) = (0); (i) < (s.size()); ++(i)) umap.insert(s[i]);
  if (n > umap.size()) {
    cout << "NO" << endl;
    return 0;
  }
  vector<int> pos;
  for (int i = 0; i < s.size() && pos.size() < n; i++) {
    if (umap.count(s[i]) == 1) {
      pos.push_back(i);
      umap.erase(s[i]);
    }
  }
  cout << "YES" << endl;
  int start = 0;
  for (int(i) = (1); (i) < (pos.size()); ++(i)) {
    cout << s.substr(start, pos[i] - start) << endl;
    start = pos[i];
  }
  cout << s.substr(start) << endl;
  return 0;
}
