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;
bool mark[100];
string answer[100];
int main() {
  int n;
  scanf("%d", &n);
  string x;
  cin >> x;
  int N = (int)x.length();
  int cont = 0;
  int li = 0;
  for (int i = 0; i < N; i++) {
    if (!mark[x[i] - 'a']) {
      if (cont + 1 == n) {
        answer[cont] = x.substr(i);
        cont++;
        break;
      }
      mark[x[i] - 'a'] = true;
      answer[cont] += x[i];
      int j;
      for (j = i + 1; mark[x[j] - 'a'] && j < N; j++) {
        answer[cont] += x[j];
      }
      i = j - 1;
      cont++;
    }
  }
  if (cont < n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < cont; i++) printf("%s\n", answer[i].c_str());
  }
  return 0;
}
