#include <bits/stdc++.h>
using namespace std;
int main() {
  int k;
  cin >> k;
  vector<int> a;
  while (k > 0) {
    a.push_back(k % 10);
    k /= 10;
  }
  int n = 2;
  int w = a.size();
  int len = 2 * w + 1;
  for (int i = 0; i < a.size(); i++) {
    n += a[i] + 10 * i + (i + 1) + (len - (2 * i + 2) - 1);
  }
  cout << n << endl;
  vector<string> g;
  string s;
  for (int i = 0; i < n + 1; i++) s.push_back('N');
  for (int i = 0; i < n + 1; i++) g.push_back(s);
  int c = 3;
  for (int i = 0; i < w; i++) {
    for (int l = 0; l < a[i]; l++) {
      g[1][c] = 'Y';
      g[c][1] = 'Y';
      c++;
    }
    for (int l = 0; l < a[i]; l++) {
      g[c][c - l - 1] = 'Y';
      g[c - l - 1][c] = 'Y';
    }
    for (int l = 0; l < i; l++) {
      for (int j = 0; j < 10; j++) {
        g[c][c + j + 1] = 'Y';
        g[c + j + 1][c] = 'Y';
      }
      c += 11;
      for (int j = 0; j < 10; j++) {
        g[c][c - j - 1] = 'Y';
        g[c - j - 1][c] = 'Y';
      }
    }
    for (int l = 0; l < (len - (2 * i + 2) - 1); l++) {
      g[c + l][c + l + 1] = 'Y';
      g[c + l + 1][c + l] = 'Y';
    }
    c += (len - (2 * i + 2) - 1);
    g[c][2] = 'Y';
    g[2][c] = 'Y';
    c++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << g[i][j];
    }
    cout << endl;
  }
  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 main() {
  int k;
  cin >> k;
  vector<int> a;
  while (k > 0) {
    a.push_back(k % 10);
    k /= 10;
  }
  int n = 2;
  int w = a.size();
  int len = 2 * w + 1;
  for (int i = 0; i < a.size(); i++) {
    n += a[i] + 10 * i + (i + 1) + (len - (2 * i + 2) - 1);
  }
  cout << n << endl;
  vector<string> g;
  string s;
  for (int i = 0; i < n + 1; i++) s.push_back('N');
  for (int i = 0; i < n + 1; i++) g.push_back(s);
  int c = 3;
  for (int i = 0; i < w; i++) {
    for (int l = 0; l < a[i]; l++) {
      g[1][c] = 'Y';
      g[c][1] = 'Y';
      c++;
    }
    for (int l = 0; l < a[i]; l++) {
      g[c][c - l - 1] = 'Y';
      g[c - l - 1][c] = 'Y';
    }
    for (int l = 0; l < i; l++) {
      for (int j = 0; j < 10; j++) {
        g[c][c + j + 1] = 'Y';
        g[c + j + 1][c] = 'Y';
      }
      c += 11;
      for (int j = 0; j < 10; j++) {
        g[c][c - j - 1] = 'Y';
        g[c - j - 1][c] = 'Y';
      }
    }
    for (int l = 0; l < (len - (2 * i + 2) - 1); l++) {
      g[c + l][c + l + 1] = 'Y';
      g[c + l + 1][c + l] = 'Y';
    }
    c += (len - (2 * i + 2) - 1);
    g[c][2] = 'Y';
    g[2][c] = 'Y';
    c++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << g[i][j];
    }
    cout << endl;
  }
  return 0;
}

```