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 CPP):

Solution:
#include <bits/stdc++.h>
char g[1001][1001];
int main() {
  int k, z, n = 2, i, j, b3[30], len = 0, from, prev, y;
  scanf("%d", &k);
  memset(g, 'N', sizeof(g));
  z = k;
  while (z) b3[len++] = z % 3, z /= 3;
  for (i = 0; i < len; i++)
    if (b3[i]) {
      from = 0;
      prev = 1;
      for (j = 0; j < i; j++) {
        for (z = 0; z < prev; z++)
          for (y = 0; y < 3; y++) g[from + z][n + y] = 'Y';
        from = n;
        prev = 3;
        n += 3;
      }
      for (z = 0; z < prev; z++)
        for (y = 0; y < b3[i]; y++) g[from + z][n + y] = 'Y';
      from = n;
      prev = b3[i];
      n += b3[i];
      for (; j < len - 1; j++) {
        for (z = 0; z < prev; z++) g[from + z][n] = 'Y';
        from = n;
        n++;
        prev = 1;
      }
      for (z = 0; z < prev; z++) g[from + z][1] = 'Y';
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      if (g[i][j] == 'Y') g[j][i] = 'Y';
  printf("%d\n", n);
  for (i = 0; i < n; i++) {
    g[i][n] = 0;
    puts(g[i]);
  }
  return 0;
}