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"}.
# http://codeforces.com/problemset/problem/544/A
def main():
    n = int(input())
    s = str(input())
    if n > len(s):
        print('NO')
        return
    elif n == 1:
        print('YES')
        print(s)
    else:
        strings = []
        letters = len(set(list(s)))
        if letters < n:
            print('NO')
            return
        curr = s[0]
        comp = s[0]
        used = {s[0]: 1}
        cant = 0
        for i in range(1, len(s)):
            if s[i] == comp:
                curr += s[i]
            else:
                check = used.get(s[i], -1)
                if check != -1:
                    curr += s[i]
                    if i == len(s) - 1:
                        strings.append(curr)
                        break
                    continue
                else:
                    strings.append(curr)
                    cant += 1
                    if cant == n:
                        strings[-1] = strings[-1] + s[i:]
                        break
                    curr = s[i]
                    comp = s[i]
                    used[comp] = 1
                    if i == len(s) - 1:
                        strings.append(curr)
                        break
        if curr != strings[-1] and cant != n:
            strings.append(curr)
        print('YES')
        for i in strings:
            print(i)
main()