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 n;
int diff = 0;
string s;
map<char, int> mp, mp2;
void start() {
  cin >> n >> s;
  for (int i = 0; i < s.size(); i++) {
    if (mp[s.at(i)] == 0) {
      diff++;
      mp[s.at(i)]++;
    }
  }
  if (n <= diff) {
    cout << "YES" << endl;
    char curr;
    for (int i = 0; i < s.size(); i++) {
      if (i == 0) {
        curr = s.at(i);
        mp2[curr]++;
        cout << curr;
      } else if (n > 1) {
        if (curr != s.at(i) && mp2[s.at(i)] == 0) {
          mp2[s.at(i)]++;
          n--;
          cout << endl;
          cout << s.at(i);
        } else
          cout << s.at(i);
        curr = s.at(i);
      } else {
        cout << s.at(i);
      }
    }
    cout << endl;
  } else
    cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  start();
  return 0;
}
