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;
int main() {
  int n;
  cin >> n;
  string S;
  cin >> S;
  int ans = 0, kapa = n;
  vector<string> V;
  int A[26] = {0};
  for (int i = 0; i < S.size(); i++) {
    if (n != 0) {
      if (A[S[i] - 'a'] == 0) {
        string K = string("") + S[i];
        A[S[i] - 'a'] = 1;
        V.push_back(K);
        n--;
      } else {
        V.back().push_back(S[i]);
      }
    } else {
      V.back().push_back(S[i]);
    }
  }
  if (V.size() == kapa) {
    cout << "YES" << endl;
    for (int i = 0; i < V.size(); i++) cout << V[i] << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
