#include <bits/stdc++.h>
using namespace std;
int mat[1000][1000];
int num[1000] = {-1};
int main() {
  int cnt;
  cin >> cnt;
  int tmp = cnt;
  int roads = 0;
  int max = 0;
  for (int i = 0; tmp > 0; i++) {
    if (tmp & 1) {
      num[roads] = i;
      roads++;
      if (max < i) {
        max = i;
      }
    }
    tmp = tmp >> 1;
  }
  if (max == 0) {
    cout << "2" << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  int total = 3;
  int pos = 2;
  mat[0][pos] = 1;
  for (int i = 0; i < max; i++) {
    mat[pos][pos + 1] = 1;
    mat[pos][pos + 2] = 1;
    mat[pos + 1][pos + 3] = 1;
    mat[pos + 2][pos + 3] = 1;
    pos += 3;
    total += 3;
  }
  mat[1][pos] = 1;
  pos++;
  int sPos = pos;
  for (int i = 0; i < 2 * max; i++) {
    mat[pos][pos + 1] = 1;
    pos++;
    total++;
  }
  mat[1][pos - 1] = 1;
  for (int i = 0; i < roads - 1; i++) {
    mat[num[i] * 3 + 2][sPos + num[i] * 2] = 1;
  }
  for (int j = 0; j < total; j++) {
    for (int i = 0; i < j; i++) {
      mat[j][i] = mat[i][j];
    }
  }
  cout << total << endl;
  for (int j = 0; j < total; j++) {
    for (int i = 0; i < total; i++) {
      if (mat[j][i] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}

### 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 mat[1000][1000];
int num[1000] = {-1};
int main() {
  int cnt;
  cin >> cnt;
  int tmp = cnt;
  int roads = 0;
  int max = 0;
  for (int i = 0; tmp > 0; i++) {
    if (tmp & 1) {
      num[roads] = i;
      roads++;
      if (max < i) {
        max = i;
      }
    }
    tmp = tmp >> 1;
  }
  if (max == 0) {
    cout << "2" << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  int total = 3;
  int pos = 2;
  mat[0][pos] = 1;
  for (int i = 0; i < max; i++) {
    mat[pos][pos + 1] = 1;
    mat[pos][pos + 2] = 1;
    mat[pos + 1][pos + 3] = 1;
    mat[pos + 2][pos + 3] = 1;
    pos += 3;
    total += 3;
  }
  mat[1][pos] = 1;
  pos++;
  int sPos = pos;
  for (int i = 0; i < 2 * max; i++) {
    mat[pos][pos + 1] = 1;
    pos++;
    total++;
  }
  mat[1][pos - 1] = 1;
  for (int i = 0; i < roads - 1; i++) {
    mat[num[i] * 3 + 2][sPos + num[i] * 2] = 1;
  }
  for (int j = 0; j < total; j++) {
    for (int i = 0; i < j; i++) {
      mat[j][i] = mat[i][j];
    }
  }
  cout << total << endl;
  for (int j = 0; j < total; j++) {
    for (int i = 0; i < total; i++) {
      if (mat[j][i] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}

```