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;
string s, tmp;
int n;
vector<string> vt;
bool arr[26];
int idx = 0;
int main() {
  cin >> n >> s;
  arr[s[0] - 'a'] = true;
  for (int i = 0; i < s.size(); i++) {
    if (!arr[s[i] - 'a']) {
      arr[s[i] - 'a'] = true;
      tmp = "";
      for (int j = idx; j < i; j++) {
        tmp += s[j];
      }
      vt.push_back(tmp);
      idx = i;
    }
  }
  tmp = "";
  for (int i = idx; i < s.size(); i++) {
    tmp += s[i];
  }
  vt.push_back(tmp);
  if (vt.size() >= n) {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << vt[i];
      if (i != n - 1) cout << endl;
    }
    for (int i = n; i < vt.size(); i++) {
      cout << vt[i];
    }
    cout << endl;
    return 0;
  }
  cout << "NO" << endl;
  return 0;
}
