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.util.*;


public class ConCho {
    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        int k = input.nextInt();
        String q = input.next();

        Vector<StringBuffer> a = new Vector<>();
        int[] check = new int[30];
        Arrays.fill(check,0);

        //System.out.printf("ahihi %s\n",q);

        for(int i = 0 ; i < q.length() ; i++){
            if(a.size() >= k){
                a.lastElement().append(q.charAt(i));
            }
            else{
                if(check[q.charAt(i) - 'a'] != 0 ){
                    a.lastElement().append(q.charAt(i));
                }
                else{
                    a.add(new StringBuffer(Character.toString(q.charAt(i))));
                    check[q.charAt(i) - 'a']++;
                }
            }
        }
        if(a.size() < k)
            System.out.printf("NO\n");
        else{
            System.out.printf("YES\n");
            for(StringBuffer x: a){
                System.out.printf("%s\n",x);
            }
        }


    }

}