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() {
  char s[150], temp[150];
  int ans[150], flag[150], n, sum = 1, m, res = 0;
  scanf("%d%s", &n, s);
  memset(flag, 0, sizeof(flag));
  for (int i = 0; i < strlen(s); i++) temp[i] = s[i];
  sort(temp, temp + strlen(s));
  for (int i = 1; i < strlen(s); i++)
    if (temp[i] != temp[i - 1]) sum++;
  if (n > sum)
    cout << "NO" << endl;
  else {
    int i = 0, j = 0, k = 0;
    cout << "YES" << endl;
    for (int i = 0; i < strlen(s) - 1; i++) {
      if (s[i] != s[i + 1] && !flag[s[i + 1] - 'a']) ans[k++] = i;
      flag[s[i] - 'a'] = 1;
    }
    if (!flag[s[strlen(s) - 1] - 'a']) ans[k] = strlen(s) - 1;
    for (int i = 0; i < strlen(s); i++) {
      if (n == 1) {
        res = i;
        break;
      }
      cout << s[i];
      if (i == ans[j]) {
        j++;
        n--;
        cout << endl;
      }
    }
    for (int j = res; j < strlen(s); j++) cout << s[j];
    cout << endl;
  }
  return 0;
}
