#include <bits/stdc++.h>
using namespace std;
long long tp[32];
bool adj[1000][1000] = {0};
inline void solve() {
  long long k;
  cin >> k;
  int vc, i, j;
  tp[0] = 1;
  for (i = 1; i <= 31; i++) {
    tp[i] = tp[i - 1] + tp[i - 1];
  }
  adj[3][1] = 1;
  adj[1][3] = 1;
  adj[1][4] = 1;
  adj[4][1] = 1;
  vc = 4;
  for (i = 2; i <= 31; i++) {
    vc++;
    adj[vc][vc - 1] = 1;
    adj[vc - 1][vc] = 1;
    adj[vc][vc - 2] = 1;
    adj[vc - 2][vc] = 1;
    vc++;
    adj[vc][vc - 3] = 1;
    adj[vc - 3][vc] = 1;
    adj[vc][vc - 2] = 1;
    adj[vc - 2][vc] = 1;
  }
  for (i = 30; i >= 0; i--) {
    if (k >= tp[i]) {
      k -= tp[i];
      j = 31 - i;
      vc++;
      adj[(i + 2) * 2][vc] = 1;
      adj[vc][(i + 2) * 2] = 1;
      while (j > 1) {
        vc++;
        adj[vc][vc - 1] = 1;
        adj[vc - 1][vc] = 1;
        j--;
      }
      adj[2][vc] = 1;
      adj[vc][2] = 1;
    }
  }
  cout << vc << endl;
  for (i = 1; i <= vc; i++) {
    for (j = 1; j <= vc; j++) {
      if (adj[i][j] == 1) {
        printf("Y");
      } else {
        printf("N");
      }
    }
    printf("\n");
  }
  printf("\n");
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long tp[32];
bool adj[1000][1000] = {0};
inline void solve() {
  long long k;
  cin >> k;
  int vc, i, j;
  tp[0] = 1;
  for (i = 1; i <= 31; i++) {
    tp[i] = tp[i - 1] + tp[i - 1];
  }
  adj[3][1] = 1;
  adj[1][3] = 1;
  adj[1][4] = 1;
  adj[4][1] = 1;
  vc = 4;
  for (i = 2; i <= 31; i++) {
    vc++;
    adj[vc][vc - 1] = 1;
    adj[vc - 1][vc] = 1;
    adj[vc][vc - 2] = 1;
    adj[vc - 2][vc] = 1;
    vc++;
    adj[vc][vc - 3] = 1;
    adj[vc - 3][vc] = 1;
    adj[vc][vc - 2] = 1;
    adj[vc - 2][vc] = 1;
  }
  for (i = 30; i >= 0; i--) {
    if (k >= tp[i]) {
      k -= tp[i];
      j = 31 - i;
      vc++;
      adj[(i + 2) * 2][vc] = 1;
      adj[vc][(i + 2) * 2] = 1;
      while (j > 1) {
        vc++;
        adj[vc][vc - 1] = 1;
        adj[vc - 1][vc] = 1;
        j--;
      }
      adj[2][vc] = 1;
      adj[vc][2] = 1;
    }
  }
  cout << vc << endl;
  for (i = 1; i <= vc; i++) {
    for (j = 1; j <= vc; j++) {
      if (adj[i][j] == 1) {
        printf("Y");
      } else {
        printf("N");
      }
    }
    printf("\n");
  }
  printf("\n");
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```