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.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;


public class A_Div2_302 {
    public static void main(String[]arg) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int K = Integer.parseInt(in.readLine());
        String q = in.readLine();
        int i = 0,k; 
        char c,cc;
        boolean is = false;
        HashSet<Character> set = new HashSet<Character>();
        ArrayList<Integer> index = new ArrayList<Integer>();
        StringBuffer sb = new StringBuffer();
        if(K == 1)
        {
            System.out.println("YES");
            System.out.println(q);
        }
        else
        {
            
            k = 1;  
            c = q.charAt(0);
            set.add(c);
            for(i = 1; i < q.length() && !is; i++)
            {
                cc = q.charAt(i);
                if(!set.contains(cc))
                {
                    set.add(cc);
                    k++;
                    index.add(i);
                    if(k == K)
                    {
                        is = true;
                        break;
                    }
                }
            }
            if(!is)
                System.out.println("NO");
            else
            {
                index.add(q.length());
                i = 0;
                for(Integer j : index)
                {
                    sb.append(q.substring(i, j)+"\n");
                    i = j;
                }
                System.out.println("YES");
                System.out.print(sb);
            }
        }
    }
}
