#include <bits/stdc++.h>
using namespace std;
int arr[100][100];
int main() {
  int k;
  cin >> k;
  int side[33], first[33], sec[33];
  for (int i = 65, j = 0; i <= 95; i++, j++) side[j] = i;
  for (int i = 3, j = 0; i <= 33; i++, j++) first[j] = i;
  for (int i = 34, j = 0; i <= 64; i++, j++) sec[j] = i;
  for (int i = 0; i < 30; i++) {
    arr[side[i]][side[i + 1]] = 1;
    arr[side[i + 1]][side[i]] = 1;
    arr[first[i]][first[i + 1]] = 1;
    arr[sec[i]][sec[i + 1]] = 1;
    arr[first[i]][sec[i + 1]] = 1;
    arr[sec[i]][first[i + 1]] = 1;
    arr[first[i + 1]][first[i]] = 1;
    arr[sec[i + 1]][sec[i]] = 1;
    arr[sec[i + 1]][first[i]] = 1;
    arr[first[i + 1]][sec[i]] = 1;
  }
  arr[1][65] = 1;
  arr[65][1] = 1;
  arr[33][2] = 1;
  arr[64][2] = 1;
  arr[2][33] = 1;
  arr[2][64] = 1;
  bitset<31> bin(k);
  for (int i = 0, j = 30; i <= 30; i++, j--)
    if (bin[i]) arr[side[j]][first[j]] = 1, arr[first[j]][side[j]] = 1;
  cout << "95\n";
  for (int i = 1; i <= 95; i++) {
    for (int j = 1; j <= 95; j++) {
      if (arr[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
  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 arr[100][100];
int main() {
  int k;
  cin >> k;
  int side[33], first[33], sec[33];
  for (int i = 65, j = 0; i <= 95; i++, j++) side[j] = i;
  for (int i = 3, j = 0; i <= 33; i++, j++) first[j] = i;
  for (int i = 34, j = 0; i <= 64; i++, j++) sec[j] = i;
  for (int i = 0; i < 30; i++) {
    arr[side[i]][side[i + 1]] = 1;
    arr[side[i + 1]][side[i]] = 1;
    arr[first[i]][first[i + 1]] = 1;
    arr[sec[i]][sec[i + 1]] = 1;
    arr[first[i]][sec[i + 1]] = 1;
    arr[sec[i]][first[i + 1]] = 1;
    arr[first[i + 1]][first[i]] = 1;
    arr[sec[i + 1]][sec[i]] = 1;
    arr[sec[i + 1]][first[i]] = 1;
    arr[first[i + 1]][sec[i]] = 1;
  }
  arr[1][65] = 1;
  arr[65][1] = 1;
  arr[33][2] = 1;
  arr[64][2] = 1;
  arr[2][33] = 1;
  arr[2][64] = 1;
  bitset<31> bin(k);
  for (int i = 0, j = 30; i <= 30; i++, j--)
    if (bin[i]) arr[side[j]][first[j]] = 1, arr[first[j]][side[j]] = 1;
  cout << "95\n";
  for (int i = 1; i <= 95; i++) {
    for (int j = 1; j <= 95; j++) {
      if (arr[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
  return 0;
}

```