#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 7;
bool gra[maxn][maxn];
int max_num_point;
void print() {
  printf("%d\n", max_num_point);
  for (int i = 1; i <= max_num_point; ++i) {
    for (int j = 1; j <= max_num_point; ++j) {
      if (gra[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
int main() {
  int k;
  scanf("%d", &k);
  memset(gra, 0, sizeof(gra));
  if (k == 1) {
    gra[1][2] = 1;
    gra[2][1] = 1;
    max_num_point = 2;
    print();
    return 0;
  }
  int h = log2(k);
  max_num_point = 3 * h + 2;
  gra[3 * h + 1][2] = 1;
  gra[2][3 * h + 1] = 1;
  gra[3 * h + 2][2] = 1;
  gra[2][3 * h + 2] = 1;
  if (k & 1) {
    gra[3 * h][2] = 1;
    gra[2][3 * h] = 1;
  }
  k /= 2;
  for (int i = h - 1; i > 0; --i) {
    gra[3 * i][3 * i + 3] = 1;
    gra[3 * i + 3][3 * i] = 1;
    gra[3 * i + 1][3 * i + 4] = 1;
    gra[3 * i + 4][3 * i + 1] = 1;
    gra[3 * i + 1][3 * i + 5] = 1;
    gra[3 * i + 5][3 * i + 1] = 1;
    gra[3 * i + 2][3 * i + 4] = 1;
    gra[3 * i + 4][3 * i + 2] = 1;
    gra[3 * i + 2][3 * i + 5] = 1;
    gra[3 * i + 5][3 * i + 2] = 1;
    if (k % 2) {
      gra[3 * i][3 * i + 4] = 1;
      gra[3 * i + 4][3 * i] = 1;
      gra[3 * i][3 * i + 5] = 1;
      gra[3 * i + 5][3 * i] = 1;
    }
    k /= 2;
  }
  gra[1][3] = 1;
  gra[3][1] = 1;
  gra[1][4] = 1;
  gra[4][1] = 1;
  gra[1][5] = 1;
  gra[5][1] = 1;
  print();
  return 0;
}

### 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>
using namespace std;
const int maxn = 1e3 + 7;
bool gra[maxn][maxn];
int max_num_point;
void print() {
  printf("%d\n", max_num_point);
  for (int i = 1; i <= max_num_point; ++i) {
    for (int j = 1; j <= max_num_point; ++j) {
      if (gra[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
int main() {
  int k;
  scanf("%d", &k);
  memset(gra, 0, sizeof(gra));
  if (k == 1) {
    gra[1][2] = 1;
    gra[2][1] = 1;
    max_num_point = 2;
    print();
    return 0;
  }
  int h = log2(k);
  max_num_point = 3 * h + 2;
  gra[3 * h + 1][2] = 1;
  gra[2][3 * h + 1] = 1;
  gra[3 * h + 2][2] = 1;
  gra[2][3 * h + 2] = 1;
  if (k & 1) {
    gra[3 * h][2] = 1;
    gra[2][3 * h] = 1;
  }
  k /= 2;
  for (int i = h - 1; i > 0; --i) {
    gra[3 * i][3 * i + 3] = 1;
    gra[3 * i + 3][3 * i] = 1;
    gra[3 * i + 1][3 * i + 4] = 1;
    gra[3 * i + 4][3 * i + 1] = 1;
    gra[3 * i + 1][3 * i + 5] = 1;
    gra[3 * i + 5][3 * i + 1] = 1;
    gra[3 * i + 2][3 * i + 4] = 1;
    gra[3 * i + 4][3 * i + 2] = 1;
    gra[3 * i + 2][3 * i + 5] = 1;
    gra[3 * i + 5][3 * i + 2] = 1;
    if (k % 2) {
      gra[3 * i][3 * i + 4] = 1;
      gra[3 * i + 4][3 * i] = 1;
      gra[3 * i][3 * i + 5] = 1;
      gra[3 * i + 5][3 * i] = 1;
    }
    k /= 2;
  }
  gra[1][3] = 1;
  gra[3][1] = 1;
  gra[1][4] = 1;
  gra[4][1] = 1;
  gra[1][5] = 1;
  gra[5][1] = 1;
  print();
  return 0;
}

```