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, i, b[1000] = {}, k = 0, l, c[1000] = {}, f = 0;
  long long s;
  scanf("%d", &n);
  string a;
  cin >> a;
  l = a.length();
  for (i = 0; i < l; i++) {
    if (b[a[i] - 97] == 0) {
      b[a[i] - 97] = 1;
      c[k++] = i;
    }
  }
  if (k >= n) {
    printf("YES\n");
    f = 1;
  } else
    printf("NO\n");
  if (f == 1) {
    i = 1;
    int j = 0;
    while (i < n) {
      while (j < c[i]) {
        printf("%c", a[j]);
        j++;
      }
      i++;
      printf("\n");
    }
    while (j < l) {
      printf("%c", a[j]);
      j++;
    }
    printf("\n");
  }
  return 0;
}
