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:
#!/usr/bin/env python


def read():
    return int(raw_input())


def cntBit(n):
    if n == 0:
        return 0
    return 1 + cntBit(n >> 1)


def work(n):
    nBit = cntBit(n)

    nNode = 2
    branchPoint = []
    edge = []

    edge.append((0, nNode))
    branchPoint.append(nNode)
    nNode += 1
    
    for i in range(nBit - 1):
        edge.append((nNode - 1, nNode    ))
        edge.append((nNode    , nNode + 2))
        edge.append((nNode - 1, nNode + 1))
        edge.append((nNode + 1, nNode + 2))
        branchPoint.append(nNode + 2)
        nNode += 3
    edge.append((nNode - 1, 1))

    
    for i in range(nBit - 1):
        if n & (1<<i):
            edge.append((branchPoint[i], nNode))
            nNode += 1
            for j in range(i, nBit - 1):
                edge.append((nNode - 1, nNode    ))
                if j == nBit - 2:
                    edge.append((nNode    , 1))
                    nNode += 1
                else:
                    edge.append((nNode    , nNode + 1))
                    nNode += 2

    
    adj = [['N' for j in range(nNode)] for i in range(nNode)]
    for e in edge:
        adj[e[0]][e[1]] = adj[e[1]][e[0]] = 'Y'

    print nNode
    for i in range(nNode):
        print "".join(adj[i])
    

if __name__ == "__main__":
    work(read())