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;
inline bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int a[200005], b[200005];
char s[200005], r[200005];
int test;
int main() {
  int t, n, m, k, l, x, y, z, flag, count, d, mx, mn;
  long long int sum, prod;
  scanf("%d%s", &k, s);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    a[s[i] - 'a']++;
  }
  count = 0;
  for (int i = 0; i < 26; i++) {
    if (a[i]) count++;
  }
  if (count >= k) {
    printf("YES\n");
    int i = 0;
    while (k > 1) {
      if (!b[s[i] - 'a']) {
        b[s[i] - 'a'] = 1;
        while (b[s[i] - 'a']) {
          printf("%c", s[i]);
          i++;
        }
      }
      printf("\n");
      k--;
    }
    for (int j = i; j < n; j++) printf("%c", s[j]);
  } else
    printf("NO\n");
  for (int i = 0; i < 26; i++) a[i] = b[i] = 0;
  return 0;
}
