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 k;
const int N = 105;
char s[N];
int cnt[30];
int vis[N];
int main() {
  while (~scanf("%d", &k)) {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= len; i++) {
      cnt[s[i] - 'a']++;
    }
    int x = 0;
    for (int i = 0; i < 26; i++) {
      if (cnt[i]) {
        x++;
      }
    }
    if (x >= k) {
      printf("YES\n");
    } else {
      printf("NO\n");
      continue;
    }
    memset(vis, 0, sizeof(vis));
    int cc = 0;
    for (int i = 1; i <= len; i++) {
      if (cc < k && !vis[s[i] - 'a']) {
        vis[s[i] - 'a'] = 1;
        if (cc != 0) {
          printf("\n");
        }
        cc++;
      }
      printf("%c", s[i]);
    }
    printf("\n");
  }
  return 0;
}
