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 in;
int k;
map<char, int> mp;
vector<string> ans(27);
int main() {
  cin >> k;
  cin >> in;
  if (in.size() < k) {
    cout << "NO";
    exit(0);
  }
  int a = 0;
  int i;
  for (i = 0; i < in.size() && a < k;) {
    int j;
    if (mp[in[i]] == 1) {
      ans[a - 1].push_back(in[i]);
      i++;
      continue;
    }
    mp[in[i]] = 1;
    ans[a].push_back(in[i]);
    for (j = i + 1; j < in.size() && in[i] == in[j]; ++j) {
      ans[a].push_back(in[j]);
    }
    i = j;
    a++;
  }
  while (i < in.size()) {
    ans[a - 1].push_back(in[i]);
    i++;
  }
  if (a == k) {
    cout << "YES\n";
    for (int i = 0; i < k; i++) {
      cout << ans[i] << endl;
    }
  } else {
    cout << "NO";
  }
}
