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 int MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN();
int main() {
  int n;
  scanf("%d", &n);
  string a;
  cin >> a;
  int b[26];
  for (int i = 0; i < 26; i++) b[i] = -1;
  vector<int> d;
  int cnt = 0;
  for (int i = 0; i < a.size(); i++) {
    if (b[a[i] - 'a'] == -1) {
      cnt++;
      b[a[i] - 'a'] = i;
      if (d.size() < n) d.push_back(i);
    }
  }
  if (cnt < n) {
    printf("NO");
  } else {
    sort(d.begin(), d.end());
    printf("YES\n");
    for (int i = 0; i < d.size() - 1; i++)
      cout << a.substr(d[i], d[i + 1] - d[i]) << endl;
    cout << a.substr(d[d.size() - 1], a.size() - d[d.size() - 1]);
  }
}
