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"}.
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

public class GFG {
	public static void main (String[] args) {
// 		System.out.println("GfG!");
        
        Scanner scn = new Scanner(System.in);
        int k = scn.nextInt();
        String str = scn.next();
        
        if(k > str.length()) {
            // System.out.println("here");
            System.out.println("NO");
            return;
        }
        if(k == 1) {
            System.out.println("YES");
            System.out.println(str);
            return;
        }
        
        int temp = k;
        
        Set<Character> hs = new HashSet<>();
        ArrayList<String> ans = new ArrayList<>();
        String curr = "";
        
        for (int i = 0; i < str.length(); i++) {
            char ch = str.charAt(i);
            if(k == 1) {
                curr = curr + str.substring(i);
                break;
            }
            
            if(i == 0) {
                hs.add(ch);
                curr += ch; 
            } else {
                if(hs.contains(ch)) {
                    curr += ch;
                } else {
                    ans.add(curr);
                    curr = ch + "";
                    k--;
                    hs.add(ch);
                }
            }
        }
        
        ans.add(curr);
        // for (int i = 0; i < ans.size(); i++) {
        //     System.out.print(ans.get(i) + " ");
        // }
        
        if(ans.size() == temp) {
            System.out.println("YES");
            for (int i = 0; i < ans.size(); i++) {
                System.out.print(ans.get(i) + " ");
            }
        } else {
            System.out.println("NO");
        }
        
	}
}