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.
import sys

k = int(sys.stdin.read().strip())

if k == 1:
    sys.stdout.write('2\nNY\nYN')
elif k == 2:
    sys.stdout.write('4\nNNYY\nNNYY\nYYNN\nYYNN')
elif k == 3:
    sys.stdout.write('5\nNNYYY\nNNYYY\nYYNNN\nYYNNN\nYYNNN')
else:
    pwr = 0
    while 2**(pwr+1) <= k:
        pwr += 1

    vert = []
    v = -1
    for i in xrange(pwr):
        if i == 0:
            vert += [(1, 3), (1, 4), (3, 5), (4, 5)]
            v = 5
        elif i == pwr - 1:
            vert += [(v, v+1), (v, v+2), (v+1, 2), (v+2, 2)]
            v += 2
        else:
            vert += [(v, v+1), (v, v+2), (v+1, v+3), (v+2, v+3)]
            v += 3
    for i in xrange(pwr):
        if k & (2 ** i):
            if i == 0:
                vstart = 1
            else:
                vstart = 2 + 3*i

            vert += [(vstart, v+1)]
            v += 1
            for j in xrange(2 * (pwr-i-1)):
                vert += [(v, v+1)]
                v += 1
            vert += [(v, 2)]

    sys.stdout.write("%d\n" % v)
    vz = [(['N']*v) for i in xrange(v)]

    for a, b in vert:
        vz[a-1][b-1] = vz[b-1][a-1] = 'Y'

    sys.stdout.write('\n'.join([''.join(i) for i in vz]))
