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.Scanner;
import java.util.LinkedList;
public class main
{
    public static void main(String args[])
    {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        input.nextLine();
        String a = input.nextLine();
        int arr[] = new int[200];
        int brr[] = new int[200];
        for(int i=0;i<a.length();i++)
        {
            arr[a.charAt(i)]++;
        }
        int count=0;
        for(int i=0;i<200;i++)
        {
            if(arr[i] != 0)
            {
                count++;
            }
        }
        if(count < n)
        {
            System.out.println("NO");
        }
        else
        {
            System.out.println("YES");
            int k = 0;
            brr[a.charAt(0)]++;
            for(int i=0;i<n-1;i++)
            {
                for(int j=k+1;j<a.length();j++)
                {
                    if(brr[a.charAt(j)] == 0)
                    {
                        System.out.println(a.substring(k,j));
                        brr[a.charAt(j)]++;
                        k = j;
                        break;
                    }
                }
            }
            System.out.println(a.substring(k,a.length()));
        }
    }
}