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;
vector<char> v[150];
map<char, int> z, r;
map<char, int>::iterator it;
string s;
int main() {
  int n, c = 0;
  cin >> n >> s;
  for (int i = 0; i < s.length(); i++) z[s[i]]++;
  for (it = z.begin(); it != z.end(); it++) {
    if (it->second != 0) c++;
  }
  if (n > c)
    cout << "NO";
  else {
    cout << "YES\n";
    int j = 0;
    v[j].push_back(s[0]);
    r[s[0]]++;
    for (int i = 1; i < s.length(); i++) {
      if (r[s[i]] == 0) {
        j++;
        v[j].push_back(s[i]);
        r[s[i]]++;
      } else {
        v[j].push_back(s[i]);
      }
    }
    int p = 0, q = 0;
    while (n > 1) {
      for (int j = 0; j < v[p].size(); j++) {
        cout << v[p][j];
        q++;
      }
      cout << endl;
      p++;
      n--;
    }
    for (int i = q; i < s.length(); i++) cout << s[i];
  }
  return 0;
}
