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, count = 0, index = 0;
  string s;
  cin >> n >> s;
  if (n == 1) {
    cout << "YES" << endl;
    cout << s << endl;
    return 0;
  } else if (s.length() < n) {
    cout << "NO" << endl;
    return 0;
  }
  string *ans = new string[n];
  int alphabets[26] = {0};
  char prev = s[index];
  alphabets[prev - 97] = 1;
  ans[count++] = prev;
  index++;
  do {
    char prev = s[index];
    if (alphabets[prev - 97] == 1) {
      ans[count - 1] += s[index];
      index++;
    } else {
      alphabets[prev - 97] = 1;
      ans[count] += prev;
      index++;
      count++;
    }
  } while (count != n && index != s.length());
  if (count < n) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    while (index <= s.length() - 1) ans[n - 1] += s[index++];
    for (int i = 0; i < n; i++) {
      cout << ans[i] << endl;
    }
  }
  return 0;
}
