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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class SetOfStrings {
	public static void main(String[] args) {
		PrintWriter pw = new PrintWriter(System.out);
	    reader(System.in);
	    int n = nI();
	    String s = n();
	    int[] arr = new int[26];
	    StringBuilder sb = new StringBuilder();
	    for (int i = 0; i < s.length(); i++) {
	    	if (arr[s.charAt(i) - 'a'] == 0 && n > 0) {
	    		arr[s.charAt(i) - 'a']++;
	    		sb.append("\n").append(s.charAt(i));
	    		n--;
	    	}
	    	else {
	    		sb.append(s.charAt(i));
	    	}
	    }
	    pw.println(n > 0 ? "NO" : "YES" + sb);
	    pw.close();
	}
	
	private static BufferedReader br;
	private static StringTokenizer st;

	static void reader(InputStream stream) {
	    try {
	        br = new BufferedReader(new InputStreamReader(stream));
	    } catch (Exception e) {
	        e.printStackTrace();
	    }
	}

	static String n() {
	    while (st == null || !st.hasMoreTokens()) {
	        try {
	            st = new StringTokenizer(br.readLine());
	        } catch (IOException e) {
	            e.printStackTrace();
	        }
	    }
	    return st.nextToken();
	}

	static int nI() {
	    return Integer.parseInt(n());
	}
	
	static long nL() {
	    return Long.parseLong(n());
	}
}