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 c[1000];
int a[1000];
int main() {
  int n;
  cin >> n;
  memset(c, 0, sizeof(c));
  memset(a, 0, sizeof(a));
  scanf("%s", c);
  for (int i = 0; c[i] != '\0'; i++) a[c[i]] = 1;
  int ans = 0;
  for (int i = 0; i <= 26; i++)
    if (a[i + 'a']) ans++;
  if (ans >= n) {
    printf("YES\n");
    for (int i = 0; c[i] != '\0';) {
      if (a[c[i]]) {
        a[c[i]] = 0;
        printf("%c", c[i]);
        i++;
        while ((!a[c[i]] || n == 1) && c[i] != '\0') {
          printf("%c", c[i]);
          i++;
        }
      }
      n--;
      cout << endl;
    }
  } else
    printf("NO\n");
  return 0;
}
