#include <bits/stdc++.h>
using namespace std;
int kra[1002][1002];
int main() {
  int n, kol = 3, last = 1, dl = 0;
  cin >> n;
  if (n == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  string s = "";
  while (n > 0) {
    if (n % 2 == 1)
      s += "1";
    else
      s += "0";
    n /= 2;
  }
  for (int i = s.size() - 2; i >= 0; i--) {
    kra[last][kol] = 1;
    kra[last][kol + 1] = 1;
    kra[kol][last] = 1;
    kra[kol + 1][last] = 1;
    last = kol + 2;
    kra[last][kol] = 1;
    kra[last][kol + 1] = 1;
    kra[kol][last] = 1;
    kra[kol + 1][last] = 1;
    kol += 3;
    dl += 2;
    if (s[i] == '1') {
      int pop = 1;
      for (int j = 0; j < dl - 1; j++) {
        kra[kol][pop] = 1;
        kra[pop][kol] = 1;
        pop = kol;
        kol++;
      }
      kra[pop][last] = 1;
      kra[last][pop] = 1;
    }
  }
  kra[last][2] = 1;
  kra[2][last] = 1;
  cout << kol - 1 << "\n";
  for (int i = 1; i < kol; i++) {
    for (int j = 1; j < kol; j++) {
      if (kra[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\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 kra[1002][1002];
int main() {
  int n, kol = 3, last = 1, dl = 0;
  cin >> n;
  if (n == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  string s = "";
  while (n > 0) {
    if (n % 2 == 1)
      s += "1";
    else
      s += "0";
    n /= 2;
  }
  for (int i = s.size() - 2; i >= 0; i--) {
    kra[last][kol] = 1;
    kra[last][kol + 1] = 1;
    kra[kol][last] = 1;
    kra[kol + 1][last] = 1;
    last = kol + 2;
    kra[last][kol] = 1;
    kra[last][kol + 1] = 1;
    kra[kol][last] = 1;
    kra[kol + 1][last] = 1;
    kol += 3;
    dl += 2;
    if (s[i] == '1') {
      int pop = 1;
      for (int j = 0; j < dl - 1; j++) {
        kra[kol][pop] = 1;
        kra[pop][kol] = 1;
        pop = kol;
        kol++;
      }
      kra[pop][last] = 1;
      kra[last][pop] = 1;
    }
  }
  kra[last][2] = 1;
  kra[2][last] = 1;
  cout << kol - 1 << "\n";
  for (int i = 1; i < kol; i++) {
    for (int j = 1; j < kol; j++) {
      if (kra[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
  return 0;
}

```