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;
long long int h[100001];
long long int r[100001];
long long int l[100001];
map<char, int> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int k;
  cin >> k;
  string s;
  cin >> s;
  set<char> ss;
  long long int n = s.size();
  for (long long int i = 0; i < n; i++) ss.insert(s[i]);
  if (ss.size() < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  long long int cnt = 1;
  long long int i = 0, last = 0;
  while (cnt <= k) {
    if (cnt == k) {
      while (i < n) {
        m[s[i]]++;
        cout << s[i++];
      }
      cout << endl;
      break;
    }
    cout << s[i];
    m[s[i++]]++;
    while (i < n && m[s[i]] > 0) {
      m[s[i]]++;
      cout << s[i++];
    }
    cout << endl;
    last = i;
    cnt++;
  }
  return 0;
}
