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;
vector<string> v;
bool letterAvailable(char c) {
  for (int i = 0; i < v.size(); i++) {
    string s = v[i];
    if (c == s[0]) return false;
  }
  return true;
}
int main() {
  int k, wordStart = 0;
  string str;
  cin >> k >> str;
  char aux = str[0];
  for (int i = 1; i < str.length() && k > 1; i++) {
    if (str[i] != aux && letterAvailable(str[i])) {
      v.push_back(str.substr(wordStart, i - wordStart));
      aux = str[i];
      wordStart = i;
      k--;
    }
  }
  if (k == 1) {
    v.push_back(str.substr(wordStart));
    cout << "YES" << endl;
    for (int i = 0; i < v.size(); i++) cout << v[i] << endl;
  } else
    cout << "NO" << endl;
  return 0;
}
