#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
const int inf = 0x3f3f3f3f;
int k;
char mat[maxn][maxn];
int main() {
  for (int i = 1; i <= 100; i++)
    for (int j = 1; j <= 100; j++) mat[i][j] = 'N';
  mat[2][11] = mat[11][2] = mat[1][41] = mat[41][1] = mat[1][71] = mat[71][1] =
      'Y';
  for (int i = 11; i < 40; i++) mat[i][i + 1] = mat[i + 1][i] = 'Y';
  for (int i = 41; i < 70; i++) {
    mat[i][i + 1] = mat[i + 1][i] = 'Y';
    mat[i][i + 30 + 1] = mat[i + 30 + 1][i] = 'Y';
  }
  for (int i = 71; i < 100; i++) {
    mat[i][i + 1] = mat[i + 1][i] = 'Y';
    mat[i][i - 30 + 1] = mat[i - 30 + 1][i] = 'Y';
  }
  scanf("%d", &k);
  int len = log2(k) + 1;
  for (int i = 1; i <= len; i++) {
    if (k % 2) mat[40 + i][11 + len - i] = mat[11 + len - i][40 + i] = 'Y';
    k /= 2;
  }
  puts("100");
  for (int i = 1; i <= 100; i++) {
    for (int j = 1; j <= 100; j++) putchar(mat[i][j]);
    puts("");
  }
  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;
const int maxn = 110;
const int inf = 0x3f3f3f3f;
int k;
char mat[maxn][maxn];
int main() {
  for (int i = 1; i <= 100; i++)
    for (int j = 1; j <= 100; j++) mat[i][j] = 'N';
  mat[2][11] = mat[11][2] = mat[1][41] = mat[41][1] = mat[1][71] = mat[71][1] =
      'Y';
  for (int i = 11; i < 40; i++) mat[i][i + 1] = mat[i + 1][i] = 'Y';
  for (int i = 41; i < 70; i++) {
    mat[i][i + 1] = mat[i + 1][i] = 'Y';
    mat[i][i + 30 + 1] = mat[i + 30 + 1][i] = 'Y';
  }
  for (int i = 71; i < 100; i++) {
    mat[i][i + 1] = mat[i + 1][i] = 'Y';
    mat[i][i - 30 + 1] = mat[i - 30 + 1][i] = 'Y';
  }
  scanf("%d", &k);
  int len = log2(k) + 1;
  for (int i = 1; i <= len; i++) {
    if (k % 2) mat[40 + i][11 + len - i] = mat[11 + len - i][40 + i] = 'Y';
    k /= 2;
  }
  puts("100");
  for (int i = 1; i <= 100; i++) {
    for (int j = 1; j <= 100; j++) putchar(mat[i][j]);
    puts("");
  }
  return 0;
}

```