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() {
  vector<int> ans;
  int a;
  char s[150];
  scanf("%d", &a);
  getchar();
  scanf("%s", s);
  int len = strlen(s);
  if (len < a) {
    printf("NO\n");
    return 0;
  }
  ans.push_back(0);
  for (int i = 1; i < len; i++) {
    int l = ans.size();
    if (l == a) break;
    bool f = true;
    for (int k = 0; k < l; k++)
      if (s[ans[k]] == s[i]) {
        f = false;
        break;
      }
    if (f) ans.push_back(i);
  }
  int len2 = ans.size();
  if (len2 < a) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (int i = 0; i < (len2 - 1); i++) {
    for (int k = ans[i]; k < ans[i + 1]; k++) printf("%c", s[k]);
    printf("\n");
  }
  for (int i = ans[len2 - 1]; i < len; i++) printf("%c", s[i]);
  printf("\n");
  ans.clear();
  return 0;
}
