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;
const long long dx[] = {1, 0, -1, 0};
const long long dy[] = {0, 1, 0, -1};
long long ans;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<vector<char> > v(n);
  int cntr = 0;
  map<char, int> mp;
  for (int i = 0; i < s.size(); i++) {
    if (i == 0) {
      v[0].push_back(s[0]);
      mp[s[i]]++;
      continue;
    }
    if (!mp[s[i]] and cntr < n - 1) {
      cntr++;
      mp[s[i]]++;
      v[cntr].push_back(s[i]);
    } else {
      v[cntr].push_back(s[i]);
    }
  }
  if (cntr == n - 1) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (auto j : v[i]) {
      cout << j;
    }
    cout << endl;
  }
  return 0;
}
