#include <bits/stdc++.h>
using namespace std;
int main() {
  long long kk;
  cin >> kk;
  vector<int> powers;
  int mx = 0;
  for (int i = 0; i < 32; i++) {
    if (((1 << i) & kk) > 0) {
      powers.push_back(i);
      mx = i;
    }
  }
  int n = 2;
  for (int i = 0; i < powers.size(); i++) n += 2 * powers[i];
  int k = 2;
  vector<string> res(n + mx + 5, string(n + mx + 5, 'N'));
  reverse(powers.begin(), powers.end());
  for (int i = 0; i < powers.size(); i++) {
    if (powers[i] == 0) {
      res[0][n + powers[i] + 1] = 'Y';
      res[n + 1 + powers[i]][0] = 'Y';
      break;
    }
    res[0][k] = res[k][0] = 'Y';
    res[0][k + 1] = res[k + 1][0] = 'Y';
    k += 2;
    for (int j = 1; j < powers[i]; j++) {
      res[k][k - 1] = res[k - 1][k] = 'Y';
      res[k][k - 2] = res[k - 2][k] = 'Y';
      res[k + 1][k - 1] = res[k - 1][k + 1] = 'Y';
      res[k + 1][k - 2] = res[k - 2][k + 1] = 'Y';
      k += 2;
    }
    res[k - 1][n + powers[i] + 1] = 'Y';
    res[n + 1 + powers[i]][k - 1] = 'Y';
    res[k - 2][n + powers[i] + 1] = 'Y';
    res[n + 1 + powers[i]][k - 2] = 'Y';
  }
  for (int i = n + 1; i + 1 < res.size(); i++) {
    res[i][i + 1] = res[i + 1][i] = 'Y';
  }
  int i = res.size() - 1;
  res[i][1] = res[1][i] = 'Y';
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) cout << res[i] << endl;
}

### Prompt

Please formulate 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 main() {
  long long kk;
  cin >> kk;
  vector<int> powers;
  int mx = 0;
  for (int i = 0; i < 32; i++) {
    if (((1 << i) & kk) > 0) {
      powers.push_back(i);
      mx = i;
    }
  }
  int n = 2;
  for (int i = 0; i < powers.size(); i++) n += 2 * powers[i];
  int k = 2;
  vector<string> res(n + mx + 5, string(n + mx + 5, 'N'));
  reverse(powers.begin(), powers.end());
  for (int i = 0; i < powers.size(); i++) {
    if (powers[i] == 0) {
      res[0][n + powers[i] + 1] = 'Y';
      res[n + 1 + powers[i]][0] = 'Y';
      break;
    }
    res[0][k] = res[k][0] = 'Y';
    res[0][k + 1] = res[k + 1][0] = 'Y';
    k += 2;
    for (int j = 1; j < powers[i]; j++) {
      res[k][k - 1] = res[k - 1][k] = 'Y';
      res[k][k - 2] = res[k - 2][k] = 'Y';
      res[k + 1][k - 1] = res[k - 1][k + 1] = 'Y';
      res[k + 1][k - 2] = res[k - 2][k + 1] = 'Y';
      k += 2;
    }
    res[k - 1][n + powers[i] + 1] = 'Y';
    res[n + 1 + powers[i]][k - 1] = 'Y';
    res[k - 2][n + powers[i] + 1] = 'Y';
    res[n + 1 + powers[i]][k - 2] = 'Y';
  }
  for (int i = n + 1; i + 1 < res.size(); i++) {
    res[i][i + 1] = res[i + 1][i] = 'Y';
  }
  int i = res.size() - 1;
  res[i][1] = res[1][i] = 'Y';
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) cout << res[i] << endl;
}

```