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 PYTHON3):

Solution:
class Node(object):
    cnt = 0
    def __init__(_, adj=None):
        _.id = Node.cnt
        Node.cnt += 1
        _.adj = adj if adj else []
        _.seen = False

n = int(input())
start = Node()
end = Node()
choose = Node()
for i in range(30):
    if n & (1<<i):
        choose.adj.append(end)
    if i == 29:
        break
    end = Node([Node([end]), Node([end])])
    choose = Node([Node([choose])])

start.adj.append(choose)
ans = [['N'] * Node.cnt for i in range(Node.cnt)]
def dfs(v):
    if v.seen:
        return
    v.seen = True
    for u in v.adj:
        ans[v.id][u.id] = ans[u.id][v.id] = 'Y'
        dfs(u)
dfs(start)

print(Node.cnt)
for line in ans:
    print(''.join(line))