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.math.BigDecimal;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        int i, j, t = 0;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //  t = Integer.parseInt(br.readLine());
        t = 1;
        while (t-- != 0) {
            int n = Integer.parseInt(br.readLine());
            i = 0;
            String[] ans = new String[n];
            String str = br.readLine();
            int[] a = new int[26];
            for(j = 0; j < 26; j++) {
                a[j] = 0;
            }
            j = 0;
            while(j < str.length()) {
//              System.out.println("init " + j + " " + i);
                int temp = j;
                a[str.charAt(j++) - 'a'] = 1;
//              System.out.println("initiate " + j + " " + i);
                while(j < str.length() && (a[str.charAt(j) - 'a'] == 1 || i == n-1)) {
                    j++;
                }
//              System.out.println("dest " + j);
                ans[i++] = str.substring(temp,j);
            }
            if(i == n) {
                System.out.println("YES");
                for(j = 0; j < i; j++) {
                    System.out.println(ans[j]);
                }
            } else {
                System.out.println("NO");
            }
        }
    }
}