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.
from collections import defaultdict
k=int(input())
mask=0
d=defaultdict(lambda:0)
while(mask<=30):

    if k&(1<<mask):
        d[mask]=1
        ma=mask

    mask+=1

adj=defaultdict(lambda:"N")
currl=1
currvu=3
prevu=[1]
prevl=[]
currvl=4
m=4
while((currl//2)<=ma):
    if d[currl//2]:
        adj[currvu,currvl]="Y"
        adj[currvl, currvu] = "Y"

    for j in prevu:
        adj[currvu, j] = "Y"
        adj[j, currvu] = "Y"

    for j in prevl:
        adj[currvl, j] = "Y"
        adj[j, currvl] = "Y"

    if ((currl+2)//2)<=ma:



        prevu=[currvl+1,currvl+2]

        for j in prevu:
            adj[currvu, j] = "Y"
            adj[j, currvu] = "Y"

        prevl=[currvl+3]
        for j in prevl:
            adj[currvl, j] = "Y"
            adj[j, currvl] = "Y"

        currvu=currvl+4
        currvl=currvu+1
        m=max(m,currvl)

    else:
        break

    currl+=2

print(m)
adj[2,currvl]="Y"
adj[currvl,2]="Y"

for i in range(1,m+1):
    for j in range(1,m+1):
        print(adj[i,j],end="")

    print()









