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() {
  cin.tie(NULL);
  ios::sync_with_stdio(false);
  set<int> S;
  long long int k, n, i, num = 0, x = 1;
  string str;
  vector<int> V;
  cin >> k;
  cin >> str;
  for (i = 0; i < str.length(); i++) {
    if (S.find(str[i]) == S.end()) {
      V.push_back(i);
      S.insert(str[i]);
    }
  }
  if (V.size() >= k) {
    cout << "YES" << endl;
    for (i = 0; i < (k - 1); i++)
      cout << str.substr(V[i], V[i + 1] - V[i]) << endl;
    cout << str.substr(V[k - 1], str.length() - V[k - 1]) << endl;
  } else
    cout << "NO" << endl;
  return 0;
}
