#include <bits/stdc++.h>
using namespace std;
int some, point = 2, tmp, st, en = 1;
char ans[1005][1005];
int main() {
  for (register int i = 0; i <= 1000; i++) {
    for (register int j = 0; j <= 1000; j++) {
      ans[i][j] = 'N';
    }
  }
  scanf("%d", &some);
  for (register int i = 1; i <= some; i *= 2) {
    tmp = point;
    for (register int j = st; j < en; j++) {
      ans[j][point] = 'Y';
      ans[point][j] = 'Y';
      point++;
    }
    for (register int j = st; j < en; j++) {
      ans[j][point] = 'Y';
      ans[point][j] = 'Y';
    }
    point++;
    st = tmp;
    en = point;
  }
  for (register int i = st + 1; i < en; i++) {
    if (((1 << (i - st - 1)) & some) != 0) {
      ans[i][1] = 'Y';
      ans[1][i] = 'Y';
    }
  }
  printf("%d\n", point);
  for (register int i = 0; i < point; i++) {
    for (register int j = 0; j < point; j++) {
      printf("%c", ans[i][j]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 some, point = 2, tmp, st, en = 1;
char ans[1005][1005];
int main() {
  for (register int i = 0; i <= 1000; i++) {
    for (register int j = 0; j <= 1000; j++) {
      ans[i][j] = 'N';
    }
  }
  scanf("%d", &some);
  for (register int i = 1; i <= some; i *= 2) {
    tmp = point;
    for (register int j = st; j < en; j++) {
      ans[j][point] = 'Y';
      ans[point][j] = 'Y';
      point++;
    }
    for (register int j = st; j < en; j++) {
      ans[j][point] = 'Y';
      ans[point][j] = 'Y';
    }
    point++;
    st = tmp;
    en = point;
  }
  for (register int i = st + 1; i < en; i++) {
    if (((1 << (i - st - 1)) & some) != 0) {
      ans[i][1] = 'Y';
      ans[1][i] = 'Y';
    }
  }
  printf("%d\n", point);
  for (register int i = 0; i < point; i++) {
    for (register int j = 0; j < point; j++) {
      printf("%c", ans[i][j]);
    }
    printf("\n");
  }
  return 0;
}

```