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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int vis[26];
int cnt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  string s;
  cin >> n;
  cin >> s;
  vector<string> v;
  string s1 = "";
  for (int i = 0; i < s.length(); i++) {
    if (!vis[s[i] - 'a']) {
      if (cnt < n) {
        if (s1 != "") {
          v.push_back(s1);
        }
        s1 = "";
        s1 += s[i];
        cnt++;
        vis[s[i] - 'a'] = 1;
      } else {
        s1 += s[i];
      }
    } else {
      s1 += s[i];
    }
  }
  v.push_back(s1);
  if (v.size() == n) {
    cout << "YES"
         << "\n";
    for (int i = 0; i < n; i++) {
      cout << v[i] << "\n";
    }
    return 0;
  }
  cout << "NO"
       << "\n";
}
