#include <bits/stdc++.h>
using namespace std;
int const N = 1 << 10;
bool e[N][N];
int main() {
  int k, d, c = 2, S = 0, T = 1;
  scanf("%d", &k);
  for (d = 30; d >= 0 && !(k & 1 << d); --d)
    ;
  e[S][c] = e[c][S] = true;
  for (int i = 0; i < d; ++i) {
    e[c][c + 1] = e[c + 1][c] = true;
    e[c][c + 2] = e[c + 2][c] = true;
    e[c + 1][c + 3] = e[c + 3][c + 1] = true;
    e[c + 2][c + 3] = e[c + 3][c + 2] = true;
    c += 3;
  }
  e[T][c] = e[c][T] = true;
  ++c;
  for (int i = 0; i < d; ++i)
    if (k & 1 << i) {
      int st = 3 * i + 2;
      e[st][c] = e[c][st] = true;
      e[c][c + 1] = e[c + 1][c] = true;
      ++c;
      for (int j = i + 1; j < d; ++j) {
        e[c][c + 1] = e[c + 1][c] = true;
        e[c + 1][c + 2] = e[c + 2][c + 1] = true;
        c += 2;
      }
      e[T][c] = e[c][T] = true;
      ++c;
    }
  printf("%d\n", c);
  for (int i = 0; i < c; ++i) {
    for (int j = 0; j < c; ++j) printf("%c", e[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
using namespace std;
int const N = 1 << 10;
bool e[N][N];
int main() {
  int k, d, c = 2, S = 0, T = 1;
  scanf("%d", &k);
  for (d = 30; d >= 0 && !(k & 1 << d); --d)
    ;
  e[S][c] = e[c][S] = true;
  for (int i = 0; i < d; ++i) {
    e[c][c + 1] = e[c + 1][c] = true;
    e[c][c + 2] = e[c + 2][c] = true;
    e[c + 1][c + 3] = e[c + 3][c + 1] = true;
    e[c + 2][c + 3] = e[c + 3][c + 2] = true;
    c += 3;
  }
  e[T][c] = e[c][T] = true;
  ++c;
  for (int i = 0; i < d; ++i)
    if (k & 1 << i) {
      int st = 3 * i + 2;
      e[st][c] = e[c][st] = true;
      e[c][c + 1] = e[c + 1][c] = true;
      ++c;
      for (int j = i + 1; j < d; ++j) {
        e[c][c + 1] = e[c + 1][c] = true;
        e[c + 1][c + 2] = e[c + 2][c + 1] = true;
        c += 2;
      }
      e[T][c] = e[c][T] = true;
      ++c;
    }
  printf("%d\n", c);
  for (int i = 0; i < c; ++i) {
    for (int j = 0; j < c; ++j) printf("%c", e[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}

```