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() {
  ios_base::sync_with_stdio(false);
  int k = 0, l = 0, i = 0, count = 0, x = -1, y = 0;
  vector<int> alpha(26, 0);
  string s;
  cin >> k >> s;
  char temp[105][105];
  for (i = 0; s[i] != '\0'; i++) {
    alpha[s[i] - 'a'] += 1;
    if (alpha[s[i] - 'a'] == 1) {
      ++x;
      y = 0;
      temp[x][y] = s[i];
    } else {
      y++;
      temp[x][y] = s[i];
    }
  }
  for (i = 0; i < 26; i++) {
    if (alpha[i] != 0) count++;
  }
  if (count >= k) {
    cout << "YES" << endl;
    for (i = 0; i < k; i++) {
      if (i != k - 1) {
        cout << temp[i] << endl;
      } else {
        int j = 0;
        for (j = i; j <= x; j++) cout << temp[j];
      }
    }
  } else {
    cout << "NO";
  }
  return 0;
}
