#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3;
bool adj[MAXN + 2][MAXN + 2];
int nodo_act;
void conecta(int a, int b) { adj[a][b] = adj[b][a] = true; }
int pide_nodo() {
  assert(nodo_act < MAXN);
  return ++nodo_act;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int k;
  cin >> k;
  int act = 1;
  int arriba = 3;
  nodo_act = 3;
  while (k > 1) {
    int uno, dos, tres;
    uno = pide_nodo();
    dos = pide_nodo();
    tres = pide_nodo();
    conecta(arriba, uno);
    conecta(act, dos);
    conecta(act, tres);
    if (k & 1) {
      conecta(act, uno);
      k--;
    }
    arriba = pide_nodo();
    conecta(uno, arriba);
    act = pide_nodo();
    conecta(dos, act);
    conecta(tres, act);
    k >>= 1;
  }
  conecta(arriba, 2);
  conecta(act, 2);
  int n = nodo_act;
  cout << n << '\n';
  for (int i = 1; i < n + 1; ++i) {
    for (int j = 1; j < n + 1; ++j) {
      if (adj[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << '\n';
  }
  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;
const int MAXN = 1e3;
bool adj[MAXN + 2][MAXN + 2];
int nodo_act;
void conecta(int a, int b) { adj[a][b] = adj[b][a] = true; }
int pide_nodo() {
  assert(nodo_act < MAXN);
  return ++nodo_act;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int k;
  cin >> k;
  int act = 1;
  int arriba = 3;
  nodo_act = 3;
  while (k > 1) {
    int uno, dos, tres;
    uno = pide_nodo();
    dos = pide_nodo();
    tres = pide_nodo();
    conecta(arriba, uno);
    conecta(act, dos);
    conecta(act, tres);
    if (k & 1) {
      conecta(act, uno);
      k--;
    }
    arriba = pide_nodo();
    conecta(uno, arriba);
    act = pide_nodo();
    conecta(dos, act);
    conecta(tres, act);
    k >>= 1;
  }
  conecta(arriba, 2);
  conecta(act, 2);
  int n = nodo_act;
  cout << n << '\n';
  for (int i = 1; i < n + 1; ++i) {
    for (int j = 1; j < n + 1; ++j) {
      if (adj[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << '\n';
  }
  return 0;
}

```