While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.

Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.

Input

The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got. 

Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.

Output

If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO". 

If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.

Examples

Input

5
aca
aba
aba
cab
bac


Output

YES
abacaba


Input

4
abc
bCb
cb1
b13


Output

NO


Input

7
aaa
aaa
aaa
aaa
aaa
aaa
aaa


Output

YES
aaaaaaaaa
INT_MAX = int(1e9 + 7)

def dfs(start_vertex, go_out):
	m = 0
	last = {}
	for key, value in go_out.items():
		last[key] = value - 1  # -1for each item from go_out
		m += value   # And total count of ways

	stack = [start_vertex]  # Init stack and put first vertex
	path = []  # Result path

	while stack:
		now = stack[-1]  # take one vertex

		if last[now] == -1:  # If all paths have already been gone:
			stack.pop()  # Delete this vertex from stack...
			path.append(now)  # And add to the result. Do not forget to reverse the result
			continue

		# Choosing nex vertex - simply with index las[now]. For this the 2-3 line.
		nex = vertexes[now][last[now]]

		last[now] -= 1  # This way is gone...
		m -= 1  # and here (total count)...

		stack.append(nex)  # TODO this vertex

	# print(m)
	return path if m == 0 else []  # If one or more ways have not been gone: path = []


n = int(input())
strings = [input() for _ in range(n)]

# print()

parts = set()

# Building vertexes and go_in:
vertexes = {}
for string in strings:
	part1 = string[:2]
	part2 = string[1:]

	parts.add(part1)
	parts.add(part2)

	try:
		vertexes[part1].append(part2)
	except Exception:
		vertexes[part1] = [part2]

# //////////////////

# And parts from vetexes:
# parts = list(vertexes.keys())
parts = list(parts)
# ///////////////////////


# print('vertexes =', vertexes)
# exit()

# Building go_in and go_out:
go_in = {i: 0 for i in parts}
go_out = go_in.copy()

for go_from, go_to in vertexes.items():
	go_out[go_from] += len(go_to)
	for i in go_to:
		go_in[i] += 1
# //////////////////////////

# print('parts =', parts)
# print('go_in =', go_in)
# print('go_out =', go_out)

# Serching for vertex with go_out > go_in and for the main answer: YES or NO:
count = 0
start_vertex = 0
for i in range(len(parts)):
	if go_out[parts[i]] - go_in[parts[i]] > 1:
		count = INT_MAX
		break
	if go_out[parts[i]] != go_in[parts[i]]:
		count += 1
		if go_out[parts[i]] > go_in[parts[i]]:
			start_vertex = i

# ////////////////////////////////////////
# print(count)
if count > 2:
	print('NO')
else:
	path = dfs(parts[start_vertex], go_out)

	# print(path)

	if path:
		print('YES')

		path.reverse()
		# print(path)
		# exit()

		# while m:
		# 	flag = False
		# 	for i in range(len(answer)):
		# 		for new, done in vertexes[answer[i]]:
		# 			if not done:
		# 				answer = answer[:i] + dfs(answer[i]) + answer[i + 1:]
		# 				flag = True
		# 				break
		# 		if flag:
		# 			break

		# Building result
		result = list(path[0])
		for i in range(1, len(path)):
			result.append(path[i][1])
		print(''.join(result))

	else:
		print('NO')
