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"}.
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class SetOfStrings {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int k = Integer.parseInt(br.readLine());
		String q = br.readLine();
		String[] s = new String[k];
		int start = 0, finish = 0;
		boolean used[] = new boolean['z'-'a'+1];
		int i;
		used[q.charAt(0)-'a'] = true;
		for (i = 0; i < k-1 && finish < q.length(); ++i) {
			while (finish < q.length() && used[q.charAt(finish)-'a']) {
				++finish;
			}
			s[i] = q.substring(start, finish);
			if (finish < q.length()) {
				used[q.charAt(finish)-'a'] = true;
				start = finish;
			}
		}
		if (finish < q.length()) {
			s[i] = q.substring(start, q.length());
			++i;
		}
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		if (i == k) {
			bw.write("YES\n");
			for (String si : s) {
				bw.write(si+"\n");
			}
		} else {
			bw.write("NO\n");
		}
		bw.flush();
	}

}
