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;
void solve() {
  int k;
  cin >> k;
  string s;
  cin >> s;
  if (s.length() < k) {
    cout << "NO\n";
    return;
  }
  if (k == 1) {
    cout << "YES\n";
    cout << s << "\n";
    return;
  }
  map<char, int> m;
  char a = s[0];
  m[a]++;
  vector<pair<char, int>> index;
  index.push_back({a, 0});
  vector<int> lens;
  map<char, int> index2;
  index2[a] = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] != a && m[s[i]] == 0) {
      long long len = index2[a];
      a = s[i];
      m[a]++;
      index2[s[i]] = i;
      len = i - len;
      index.push_back({s[i], i});
      lens.push_back(len);
    }
  }
  if (m.size() < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  int ct = 0;
  int j = 0;
  auto it = index.begin();
  while (it != index.end()) {
    string res = s.substr(it->second, lens[j]);
    ct++;
    cout << res << "\n";
    j++;
    if (ct == k - 1) {
      it++;
      cout << s.substr(it->second, s.length() - it->second + 1) << "\n";
      return;
    }
    it++;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}
