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>
const int INF = 500000001;
const double EPS = 1e-9;
const double PI = acos(-1.0);
using namespace std;
int main() {
  int k;
  char ch[1000];
  int f[100];
  int t[100];
  while (~scanf("%d", &k)) {
    memset(f, 0, sizeof(f));
    scanf("%s", ch);
    int len = strlen(ch);
    for (int i = 0; i < len; i++) {
      f[ch[i] - 'a'] = 1;
    }
    int res = 0;
    for (int i = 0; i < 26; i++) res += f[i];
    memset(t, 0, sizeof(t));
    if (res < k) {
      puts("NO");
    } else {
      puts("YES");
      for (int i = 0; i < len; i++) {
        if (f[ch[i] - 'a']) {
          f[ch[i] - 'a'] = 0;
          t[i] = 1;
        }
      }
      int flag = 0;
      for (int i = 0; i < len; i++) {
        if (t[i] && i && flag < k - 1) {
          flag++;
          puts("");
        }
        printf("%c", ch[i]);
      }
      puts("");
    }
  }
  return 0;
}
