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;
char str[105];
bool have[26];
int main() {
  memset(have, false, sizeof(have));
  int k;
  cin >> k;
  cin >> str;
  int length = strlen(str);
  if (length < k) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < length; i++) {
    have[str[i] - 'a'] = true;
  }
  int cnt = 0;
  for (int i = 0; i < 26; i++) {
    if (have[i]) {
      cnt++;
    }
  }
  if (cnt < k) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    have[str[0] - 'a'] = false;
    int t = 1;
    for (int i = 0; i < length; i++) {
      if (t == k) {
        for (; i < length; i++) {
          cout << str[i];
        }
        cout << endl;
        break;
      }
      cout << str[i];
      if (i == length - 1 || have[str[i + 1] - 'a'] == true) {
        cout << endl;
        t++;
        if (i < length - 1) have[str[i + 1] - 'a'] = false;
      }
    }
  }
  return 0;
}
