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 MaxN = 1000;
using namespace std;
char s[MaxN + 5];
bool vis[26];
int seq[30];
int n;
int main() {
  while (~scanf("%d", &n)) {
    scanf("%s", s);
    int len = strlen(s);
    int tot = 0;
    for (int i = 0; i < len; i++) {
      if (!vis[s[i] - 'a']) {
        seq[++tot] = i;
        vis[s[i] - 'a'] = 1;
      }
    }
    if (tot < n)
      printf("NO\n");
    else {
      seq[tot + 1] = n;
      printf("YES\n");
      for (int i = 1; i <= n; i++) {
        int end;
        if (i == n)
          end = len;
        else
          end = seq[i + 1];
        for (int j = seq[i]; j < end; j++) printf("%c", s[j]);
        printf("\n");
      }
    }
  }
  return 0;
}
