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;
const int INF = 1000001000;
const long long INFll = 1ll * INF * INF;
const long double ldINF = 1e+018;
const long double EPS = 0.000000001;
const int sdx[4] = {-1, 0, 1, 0};
const int sdy[4] = {0, 1, 0, -1};
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, pair<T, T>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  int k;
  cin >> k;
  string s;
  cin >> s;
  vector<int> pos;
  map<char, bool> used;
  int cnt = 0;
  for (int i = 0; i < s.size() && cnt < k; i++)
    if (!used[s[i]]) cnt++, used[s[i]] = true, pos.push_back(i);
  if (cnt < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  pos.push_back(s.size());
  for (int i = 0; i < pos.size() - 1; i++)
    cout << s.substr(pos[i], pos[i + 1] - pos[i]) << endl;
  return 0;
}
