#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  vector<vector<bool> > v(1000, vector<bool>(1000, false));
  int s;
  cin >> s;
  long long maxim = 0, num = 9;
  while (num < s) {
    maxim++;
    num *= 10;
  }
  int n = 2;
  vector<int> pots(10);
  pots[0] = 0;
  for (int i = 1; i <= maxim; i++) {
    int pre = pots[i - 1];
    int post = n;
    pots[i] = post;
    ++n;
    for (int j = 0; j < 10; j++) {
      v[pre][n] = true;
      v[n][pre] = true;
      v[post][n] = true;
      v[n][post] = true;
      ++n;
    }
  }
  int pot = 0;
  while (s) {
    int num = s % 10;
    s /= 10;
    int ant = pots[pot];
    for (int j = 0; j < 18 - 2 * pot; j++) {
      v[ant][n] = true;
      v[n][ant] = true;
      ant = n;
      ++n;
    }
    for (int j = 0; j < num; j++) {
      v[ant][n] = true;
      v[n][ant] = true;
      v[n][1] = true;
      v[1][n] = true;
      ++n;
    }
    pot++;
  }
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (v[i][j])
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
}

### 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 main() {
  ios_base::sync_with_stdio(false);
  vector<vector<bool> > v(1000, vector<bool>(1000, false));
  int s;
  cin >> s;
  long long maxim = 0, num = 9;
  while (num < s) {
    maxim++;
    num *= 10;
  }
  int n = 2;
  vector<int> pots(10);
  pots[0] = 0;
  for (int i = 1; i <= maxim; i++) {
    int pre = pots[i - 1];
    int post = n;
    pots[i] = post;
    ++n;
    for (int j = 0; j < 10; j++) {
      v[pre][n] = true;
      v[n][pre] = true;
      v[post][n] = true;
      v[n][post] = true;
      ++n;
    }
  }
  int pot = 0;
  while (s) {
    int num = s % 10;
    s /= 10;
    int ant = pots[pot];
    for (int j = 0; j < 18 - 2 * pot; j++) {
      v[ant][n] = true;
      v[n][ant] = true;
      ant = n;
      ++n;
    }
    for (int j = 0; j < num; j++) {
      v[ant][n] = true;
      v[n][ant] = true;
      v[n][1] = true;
      v[1][n] = true;
      ++n;
    }
    pot++;
  }
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (v[i][j])
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
}

```