#include <bits/stdc++.h>
using namespace std;
int ch[1005][1005];
int top = 2;
int buildroad(int x, int lev) {
  for (int i = 1; i <= 4; i++) top++, ch[1][top] = ch[top][1] = 1;
  for (int k = 1; k <= x; k++) {
    for (int i = 1; i <= 4; i++)
      for (int j = 1; j <= 4; j++)
        ch[top - i + 1][top + j] = ch[top + j][top - i + 1] = i;
    top += 4;
  }
  for (int k = x + 1; k <= lev; k++) {
    for (int i = 1; i <= 4; i++)
      ch[top - i + 1][top + i] = ch[top + i][top - i + 1] = 1;
    top += 4;
  }
  return top;
}
int main() {
  int k;
  scanf("%d", &k);
  const int lev = (31 - __builtin_clz(k)) / 2;
  for (int i = 0; k; i++, k >>= 2) {
    if (k & 3) {
      int x = buildroad(i, lev);
      if (k & 1) ch[x][2] = ch[2][x] = 1;
      if (k & 2) ch[x - 1][2] = ch[2][x - 1] = ch[x - 2][2] = ch[2][x - 2] = 1;
    }
  }
  printf("%d\n", top);
  for (int i = 1; i <= top; i++) {
    for (int j = 1; j <= top; j++) putchar(ch[i][j] ? 'Y' : 'N');
    putchar('\n');
  }
}

### Prompt

Generate a Cpp solution to 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;
int ch[1005][1005];
int top = 2;
int buildroad(int x, int lev) {
  for (int i = 1; i <= 4; i++) top++, ch[1][top] = ch[top][1] = 1;
  for (int k = 1; k <= x; k++) {
    for (int i = 1; i <= 4; i++)
      for (int j = 1; j <= 4; j++)
        ch[top - i + 1][top + j] = ch[top + j][top - i + 1] = i;
    top += 4;
  }
  for (int k = x + 1; k <= lev; k++) {
    for (int i = 1; i <= 4; i++)
      ch[top - i + 1][top + i] = ch[top + i][top - i + 1] = 1;
    top += 4;
  }
  return top;
}
int main() {
  int k;
  scanf("%d", &k);
  const int lev = (31 - __builtin_clz(k)) / 2;
  for (int i = 0; k; i++, k >>= 2) {
    if (k & 3) {
      int x = buildroad(i, lev);
      if (k & 1) ch[x][2] = ch[2][x] = 1;
      if (k & 2) ch[x - 1][2] = ch[2][x - 1] = ch[x - 2][2] = ch[2][x - 2] = 1;
    }
  }
  printf("%d\n", top);
  for (int i = 1; i <= top; i++) {
    for (int j = 1; j <= top; j++) putchar(ch[i][j] ? 'Y' : 'N');
    putchar('\n');
  }
}

```