Problem: Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."

Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?

Input

The first line contains a single integer k (1 ≤ k ≤ 109).

Output

You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.

The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them. 

Examples

Input

2

Output

4
NNYY
NNYY
YYNN
YYNN

Input

9

Output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

Input

1

Output

2
NY
YN

Note

In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.

In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.

Solution (in PYTHON):

Solution:
K = int(raw_input())

N = 2
res = [['N'] * 1000 for i in range(1000)]

def new_vertex():
	global N
	N += 1
	return N

def set_edge(a, b):
	global res
	res[a - 1][b - 1] = res[b - 1][a - 1] = 'Y'

def make_chain(cnt, to):
	# (1) -> (to)
	assert(cnt >= 0)
	prev = 1
	for i in range(cnt):
		new = new_vertex()
		set_edge(prev, new)
		prev = new
	set_edge(prev, to)

def link_full(a_list, b_list):
	for a in a_list:
		for b in b_list:
			set_edge(a, b)

def make_digit(digit, p10):
	# (<ret>) -> (2)
	# (digit * 10 ** p10) paths
	ret = new_vertex()
	prev = [new_vertex() for i in range(digit)]
	link_full([ret], prev)
	for i in range(p10):
		new = [new_vertex() for i in range(10)]
		link_full(prev, new)
		prev = new
	link_full(prev, [2])
	return ret

maxp = 1
while 10 ** maxp <= K:
	maxp += 1

for i in range(maxp):
	digit = (K % (10 ** (i + 1))) / (10 ** i)
	tmp = make_digit(digit, i)
	make_chain(maxp - i - 1, tmp)

print N
for i in range(N):
	print ''.join(res[i][0:N])