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"}.
'''def contains(vec, c):
    for i in range(len(vec)):
        if vec[i] == c:
            return True
    return False
'''


def different_index_starts_at(start_index, start_letters, _str):
    for index in range(start_index, len(_str)):
        # if not contains(start_letters, item):
        if _str[index] not in start_letters:
            return index

    return -1


k = int(input())
q = input()

starting_letters = [q[0]]

answer = []

start = 0
for i in range(k):
    different = different_index_starts_at(start + 1, starting_letters, q)

    if different != -1 and k != i + 1:
        starting_letters.append(q[different])
        answer.append(q[start:different])
        start = different
    else:
        answer.append(q[start:])
        break

if len(answer) == k:
    print("YES")
    for s in answer:
        print(s)
else:
    print("NO")
