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 ≤ 10^9).
-----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 G_{ij} 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: G_{ii} = 'N' and G_{ij} = G_{ji} 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.
N = int(input())
b = bin(N)[2:][::-1]
count = len(b)-1

def print_matrix(m):
    print(len(m))
    for i in range(len(m)):
        print("".join(m[i]))

m = [['N' for i in range(2*(len(b))+count)] for j in range(2*(len(b))+count)]
for i in range(count-1):
    val = len(m)-count+i
    m[val][val+1] = 'Y'
    m[val+1][val] = 'Y'

if count>0:
    m[len(m)-1][1] = 'Y'
    m[1][len(m)-1] = 'Y'

#print_matrix(m)
c = 0
c2 = len(m)-count
for i in range(0, len(m)-count, 2):
    if i >= len(m)-2-count and i != 0:
        m[i][1] = 'Y'
        m[i+1][1] = 'Y'
        m[1][i+1] = 'Y'
        m[1][i] = 'Y'
    elif i < len(m)-2-count:
        m[i][i+2] = 'Y'
        m[i][i+3] = 'Y'
        m[i+2][i] = 'Y'
        m[i+3][i] = 'Y'
        if i != 0:
       	    m[i+1][i+2] = 'Y'
       	    m[i+1][i+3] = 'Y'
       	    m[i+2][i+1] = 'Y'
       	    m[i+3][i+1] = 'Y'
    if b[c] =='1':
        if i == len(m)-count-2:
            c2 = 1
        #print_matrix(m)
        m[i][c2] = 'Y'
        m[c2][i] = 'Y'
        if i != 0:
            m[i+1][c2] = 'Y'
            m[c2][i+1] = 'Y'
    c2+=1
        #print_matrix(m)
    c+=1

print(len(m))
for i in range(len(m)):
    print("".join(m[i]))


N = int(input())
b = bin(N)[2:][::-1]
count = len(b)-1

def print_matrix(m):
    print(len(m))
    for i in range(len(m)):
        print("".join(m[i]))

m = [['N' for i in range(2*(len(b))+count)] for j in range(2*(len(b))+count)]
for i in range(count-1):
    val = len(m)-count+i
    m[val][val+1] = 'Y'
    m[val+1][val] = 'Y'

if count>0:
    m[len(m)-1][1] = 'Y'
    m[1][len(m)-1] = 'Y'

#print_matrix(m)
c = 0
c2 = len(m)-count
for i in range(0, len(m)-count, 2):
    if i >= len(m)-2-count and i != 0:
        m[i][1] = 'Y'
        m[i+1][1] = 'Y'
        m[1][i+1] = 'Y'
        m[1][i] = 'Y'
    elif i < len(m)-2-count:
        m[i][i+2] = 'Y'
        m[i][i+3] = 'Y'
        m[i+2][i] = 'Y'
        m[i+3][i] = 'Y'
        if i != 0:
       	    m[i+1][i+2] = 'Y'
       	    m[i+1][i+3] = 'Y'
       	    m[i+2][i+1] = 'Y'
       	    m[i+3][i+1] = 'Y'
    if b[c] =='1':
        if i == len(m)-count-2:
            c2 = 1
        #print_matrix(m)
        m[i][c2] = 'Y'
        m[c2][i] = 'Y'
        if i != 0:
            m[i+1][c2] = 'Y'
            m[c2][i+1] = 'Y'
    c2+=1
        #print_matrix(m)
    c+=1

print(len(m))
for i in range(len(m)):
    print("".join(m[i]))

