#include <bits/stdc++.h>
using namespace std;
int k;
int adj[1010][1010], n = 2;
int it[31];
int main() {
  int i, j, ij = 3, temp, last = -1, cur, tf = 0, z;
  queue<int> prev;
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY\nYN" << endl;
    return 0;
  }
  ij = 3;
  for (i = 0, temp = 1; temp * 2 <= k; i++, temp *= 2) it[i] = ij, ij += 2;
  ij = 3;
  last = i;
  adj[1][ij] = adj[ij][1] = 1;
  adj[1][ij + 1] = adj[ij + 1][1] = 1;
  for (; i > 1; i--) {
    adj[ij][ij + 2] = adj[ij + 2][ij] = 1;
    adj[ij][ij + 3] = adj[ij + 3][ij] = 1;
    adj[ij + 1][ij + 2] = adj[ij + 2][ij + 1] = 1;
    adj[ij + 1][ij + 3] = adj[ij + 3][ij + 1] = 1;
    ij += 2;
  }
  k -= temp;
  adj[ij][2] = adj[2][ij] = 1;
  adj[ij + 1][2] = adj[2][ij + 1] = 1;
  ij += 2;
  while (k > 0) {
    for (i = 0, temp = 1; temp * 2 <= k; i++, temp *= 2)
      ;
    k -= temp;
    if (i == last - 1) {
      i--;
      for (j = 0; j < 2; j++) {
        adj[it[i]][ij] = adj[ij][it[i]] = 1;
        for (z = i + 1; z < last - 1; z++) {
          adj[ij][ij + 1] = adj[ij + 1][ij] = 1;
          ij++;
        }
        adj[ij][2] = adj[2][ij] = 1;
        ij++;
      }
    } else {
      adj[it[i]][ij] = adj[ij][it[i]] = 1;
      for (z = i + 1; z < last - 1; z++) {
        adj[ij][ij + 1] = adj[ij + 1][ij] = 1;
        ij++;
      }
      adj[ij][2] = adj[2][ij] = 1;
      ij++;
    }
  }
  cout << ij - 1 << endl;
  for (i = 1; i < ij; i++) {
    for (j = 1; j < ij; j++) {
      if (adj[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Generate 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 k;
int adj[1010][1010], n = 2;
int it[31];
int main() {
  int i, j, ij = 3, temp, last = -1, cur, tf = 0, z;
  queue<int> prev;
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY\nYN" << endl;
    return 0;
  }
  ij = 3;
  for (i = 0, temp = 1; temp * 2 <= k; i++, temp *= 2) it[i] = ij, ij += 2;
  ij = 3;
  last = i;
  adj[1][ij] = adj[ij][1] = 1;
  adj[1][ij + 1] = adj[ij + 1][1] = 1;
  for (; i > 1; i--) {
    adj[ij][ij + 2] = adj[ij + 2][ij] = 1;
    adj[ij][ij + 3] = adj[ij + 3][ij] = 1;
    adj[ij + 1][ij + 2] = adj[ij + 2][ij + 1] = 1;
    adj[ij + 1][ij + 3] = adj[ij + 3][ij + 1] = 1;
    ij += 2;
  }
  k -= temp;
  adj[ij][2] = adj[2][ij] = 1;
  adj[ij + 1][2] = adj[2][ij + 1] = 1;
  ij += 2;
  while (k > 0) {
    for (i = 0, temp = 1; temp * 2 <= k; i++, temp *= 2)
      ;
    k -= temp;
    if (i == last - 1) {
      i--;
      for (j = 0; j < 2; j++) {
        adj[it[i]][ij] = adj[ij][it[i]] = 1;
        for (z = i + 1; z < last - 1; z++) {
          adj[ij][ij + 1] = adj[ij + 1][ij] = 1;
          ij++;
        }
        adj[ij][2] = adj[2][ij] = 1;
        ij++;
      }
    } else {
      adj[it[i]][ij] = adj[ij][it[i]] = 1;
      for (z = i + 1; z < last - 1; z++) {
        adj[ij][ij + 1] = adj[ij + 1][ij] = 1;
        ij++;
      }
      adj[ij][2] = adj[2][ij] = 1;
      ij++;
    }
  }
  cout << ij - 1 << endl;
  for (i = 1; i < ij; i++) {
    for (j = 1; j < ij; j++) {
      if (adj[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}

```