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() {
  int n;
  cin >> n;
  int nn = n;
  string s;
  cin >> s;
  vector<string> vs;
  string temps;
  temps = s[0];
  set<char> setc;
  setc.insert(s[0]);
  n--;
  for (int i = 1; i < s.size(); i++) {
    if (n > 0) {
      if (setc.count(s[i]) == 0) {
        vs.push_back(temps);
        temps = s[i];
        setc.insert(s[i]);
        n--;
      } else {
        temps += s[i];
      }
    } else
      temps += s[i];
  }
  vs.push_back(temps);
  if (vs.size() < nn) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int i = nn; i < vs.size(); i++) {
      vs[nn - 1] += vs[i];
    }
    for (int i = 0; i < nn; i++) {
      cout << vs[i] << endl;
    }
  }
  return 0;
}
