#include <bits/stdc++.h>
int ma[1005][1005] = {0};
int main() {
  int k, x, y, i, j, n = 64;
  scanf("%d", &k);
  ma[0][2] = ma[0][3] = 1;
  for (i = 4; i < 64; i += 2)
    ma[i - 2][i] = ma[i - 2][i + 1] = ma[i - 1][i] = ma[i - 1][i + 1] = 1;
  for (i = 0; i < 31; i++)
    if (k & (1 << i)) {
      x = i * 2 + 2;
      for (j = 0; j < 31 - i; j++) {
        ma[x][n] = ma[n][x] = 1;
        x = n++;
      }
      ma[1][x] = ma[x][1] = 1;
    }
  printf("%d\n", n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      if (ma[i][j] || ma[j][i])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
}

### Prompt

Create a solution in Cpp for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
int ma[1005][1005] = {0};
int main() {
  int k, x, y, i, j, n = 64;
  scanf("%d", &k);
  ma[0][2] = ma[0][3] = 1;
  for (i = 4; i < 64; i += 2)
    ma[i - 2][i] = ma[i - 2][i + 1] = ma[i - 1][i] = ma[i - 1][i + 1] = 1;
  for (i = 0; i < 31; i++)
    if (k & (1 << i)) {
      x = i * 2 + 2;
      for (j = 0; j < 31 - i; j++) {
        ma[x][n] = ma[n][x] = 1;
        x = n++;
      }
      ma[1][x] = ma[x][1] = 1;
    }
  printf("%d\n", n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      if (ma[i][j] || ma[j][i])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
}

```