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 cur, k, oz, w, l, r, i, j, k1, k2, dr, n, m, t, qt, qh, md, a[5005],
    used[5005], x, y;
string s;
vector<string> dno;
int main() {
  cin >> n >> s;
  l = 0;
  for (i = 0; i < s.size();)
    if (!used[s[i]]) {
      used[s[i]] = 1;
      dno.push_back("");
      while (used[s[i]] && i < s.size()) {
        dno[l].push_back(s[i]);
        i++;
      }
      l++;
    }
  if (l < n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (i = 0; i < n - 1; i++) cout << dno[i] << "\n";
  for (i = n - 1; i < l; i++) cout << dno[i];
  cout << endl;
  return 0;
}
