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"}.
n = int(input())
s = input()
if n == 1:
    print("YES")
    print(s)
    exit()
a = []
b = []
for i in range(26):
    a.append(0)
    b.append(-1)
for i in range(len(s)):
    a[ord(s[i])-ord('a')] += 1
    if b[ord(s[i])-ord('a')] == -1:
        b[ord(s[i])-ord('a')] = i
    #print(s[i], i)
num = 0
c = []
for i in range(26):
    if a[i] != 0:
        num += 1
        c.append(b[i])
#print(c)
i = 0
lastNum = 0
if num < n:
    print("NO")
else:
    print("YES")
    minNum = 100000
    for i in range(len(c)):
        if minNum > c[i]:
            minNum = c[i]
            minNumPos = i
    if c[minNumPos] != 0:
        print(s[:c[minNumPos]])
        lastNum = c[minNumPos]
        n -= 1
    c[minNumPos] = 100000
    while n > 1:
        minNum = 100000
        for i in range(len(c)):
            if minNum > c[i]:
                minNum = c[i]
                minNumPos = i
        print(s[lastNum:c[minNumPos]])
        lastNum = c[minNumPos]
        c[minNumPos] = 100000
        #print(c, lastNum)
        n -= 1
    print(s[lastNum:])
    #print(c, lastNum)
#if num >= n:


#print(a, b)