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[200];
int t[200], vis[200];
int main() {
  int k;
  scanf("%d%s", &k, str);
  int len = strlen(str);
  if (k > len) {
    printf("NO\n");
    return 0;
  }
  int sum = 0, num = 1;
  for (int i = 1; i <= len; i++) t[str[i - 1]] = 1;
  for (int i = 'a'; i <= 'z'; i++)
    if (t[i]) sum++;
  vis[str[0]] = 1;
  if (sum >= k) {
    printf("YES\n");
    for (int i = 1; i <= len; i++) {
      if (t[str[i - 1]] && !vis[str[i - 1]] && num < k && i != 1)
        printf("\n"), vis[str[i - 1]] = 1, num++;
      printf("%c", str[i - 1]);
    }
  } else {
    printf("NO\n");
  }
  return 0;
}
