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 i, j, k, n, m, a, b, c;
string s, str;
vector<int> V;
int freq[113];
int used[113];
int part(0);
int main(void) {
  cin >> k;
  cin >> s;
  n = s.size();
  int pivot = 0;
  for (int i = 0; i < n; ++i)
    if (freq[s[i] - 'a'] == 0) ++part, ++freq[s[i] - 'a'], ++used[i];
  if (part < k) return cout << "NO", 0;
  cout << "YES\n";
  for (i = 0; i < n && k > 0;) {
    if (used[i] == 1 && i < n) cout << s[i], ++i;
    while (used[i] != 1 && i < n) cout << s[i], ++i;
    --k;
    if (k) cout << "\n";
  }
  if (i < s.size())
    for (i; i < s.size(); ++i) cout << s[i];
  return 0;
}
