You are given a string *q*. A sequence of *k* strings *s*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*k*</sub> is called beautiful, if the concatenation of these strings is string *q* (formally, *s*<sub class="lower-index">1</sub> + *s*<sub class="lower-index">2</sub> + ... + *s*<sub class="lower-index">*k*</sub> = *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.
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.
from collections import defaultdict
def cutter(a, cuts):
    
    if cuts == 1:
        return [a]
    comparador = set([a[0]])
    aux = a[0]
    ans = []
    flag = True
    for i in range(1, len(a)):
        # print(a[i])
        if a[i] not in comparador:
            ans.append(aux)
            aux = a[i]
            comparador.union(set([a[i]]))
        else:
            aux += a[i]
        
        # print(aux)
        
        if len(ans) == cuts - 1 :
            ans.append(a[i:])
            break
    return ans
        
def isBeautiful(a):
    mapa = defaultdict(int)
    
    for i in a:
        mapa[i[0]] += 1
        
        if mapa[i[0]] > 1:
            return False
    return True

def func():
    n = int(input())
    p = input()
    
    if len(p) < n:
        print('NO')
        return
    
    lista =  cutter(p, n)
    # print(lista)
    if len(lista) < n:
        print('NO')
        return
    
    if isBeautiful(lista):
        print('YES')
        for i in lista:
            print(i)
    else:
        print('NO')
        return
    
func()