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.*;

public class Main{

static final InputStream in=System.in;
static final PrintWriter out=new PrintWriter(System.out);

public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(in));
String line="";
lo:
while((line=br.readLine())!=null&&!line.isEmpty()){
int n=Integer.parseInt(line);
line=br.readLine();
char[] q=line.toCharArray();
char[] rem=new char[n];
rem[0]=q[0];
int[] stnum=new int[n];
stnum[0]=0;
if(q.length<n){
out.println("NO");
break lo;
}
for(int i=1;i<n;i++){
lo2:
for(int j=stnum[i-1]+1;j<q.length;j++){
for(int k=0;k<i;k++){
if(q[j]==rem[k]){
if(j==q.length-1){
out.println("NO");
break lo;
}
continue lo2;
}
}
rem[i]=q[j];
stnum[i]=j;
break;
}
}
out.println("YES");
for(int i=0;i<n;i++){
if(i==n-1) out.println(line.substring(stnum[n-1]));
else out.println(line.substring(stnum[i],stnum[i+1]));
}
}
out.flush();
}
}