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"}.
#!/usr/bim/python

n = input()
s = raw_input()
j = 0
fl = 0
count = 0
L=[]
L1=[]
L2=[]
for i in s:
	if i not in L:
		L.append(i)
	
if len(L) < n:
	print "NO"

else:
	if fl == 0:
		print "YES"
		fl = 1
	while count < n-1 and j < len(s)-1:
		if s[j] == s[j+1] or s[j+1] in L1:
			#print s[j],
			L2.append(s[j])
			L1.append(s[j])
			j = j+1
		elif s[j] != s[j+1] and s[j+1] not in L1:
			#print s[j]
			L1.append(s[j])
			L2.append(s[j])	
			L2.append("\n")
			count =count +1
			j = j+1


	while j != len(s):
		#print s[j],
		L2.append(s[j])
		j = j+1
		
	print "".join(L2)
