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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
template <class T>
T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
void solve() {
  string s;
  int k;
  cin >> k;
  cin >> s;
  map<char, int> cnt;
  vector<bool> used(26);
  for (int i = 0; i < s.length(); i++) cnt[s[i]]++;
  if (cnt.size() < k) {
    cout << "NO";
    return;
  }
  cout << "YES" << endl;
  int ptr = 0;
  while (k && ptr < s.length()) {
    used[s[ptr] - 'a'] = true;
    cout << s[ptr++];
    if (ptr == s.length()) return;
    if (!used[s[ptr] - 'a']) {
      --k;
      if (k) cout << endl;
    }
  }
  while (ptr < s.length()) cout << s[ptr++];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
