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.*;
import java.io.*;
public class cadenasbonitas{
public static void main(String args[]) throws IOException{
BufferedReader lector = new BufferedReader(new InputStreamReader(System.in));
int a = Integer.parseInt(lector.readLine());
String tmp = lector.readLine();
boolean us[] = new boolean[26];
Vector<String> tal = new Vector<String>();
us[tmp.charAt(0)-'a']=true;
if(a>1)
tal.add(tmp.charAt(0)+"");
else
tal.add(tmp);
int p = 1;
for(int n= tal.get(0).length();n<tmp.length();n++){
if(us[tmp.charAt(n)-'a']){
String tt = tal.get(tal.size()-1);
tal.remove(tal.size()-1);
tal.add(tt+tmp.charAt(n));
continue;
}
tal.add(tmp.charAt(n)+"");
us[tmp.charAt(n)-'a']=true;
p++;
if(p==a){
String aa = tal.get(tal.size()-1);
tal.remove(tal.size()-1);
tal.add(aa+tmp.substring(n+1));
break;
}
}
System.out.println((tal.size()!=a)?"NO":"YES");
if(tal.size()!=a)return;
for(int n = 0;n<tal.size();n++)System.out.println(tal.get(n));
}
}
