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 s[150];
int main() {
  int n;
  int b[30];
  memset(b, -1, sizeof(b));
  scanf("%d", &n);
  getchar();
  scanf("%s", s);
  int ans = 0;
  for (int i = 0; i < strlen(s); i++)
    if (b[s[i] - 'a'] == -1) {
      ans++;
      b[s[i] - 'a'] = i;
    }
  int k = 0;
  int c[30];
  int vis[30];
  memset(vis, 0, sizeof(vis));
  if (ans >= n) {
    printf("YES\n");
    for (int i = 0; i < strlen(s); i++) {
      if (b[s[i] - 'a'] != -1 && vis[s[i] - 'a'] == 0) {
        c[k++] = b[s[i] - 'a'];
        vis[s[i] - 'a'] = 1;
      }
      if (k == n) break;
    }
    for (int i = 0; i < n - 1; i++) {
      for (int j = c[i]; j < c[i + 1]; j++) printf("%c", s[j]);
      printf("\n");
    }
    for (int j = c[n - 1]; j < strlen(s); j++) printf("%c", s[j]);
    printf("\n");
  } else
    printf("NO\n");
  return 0;
}
