#include <bits/stdc++.h>
using namespace std;
int N, K;
bool is[1002][1002];
int wh[52];
int main() {
  cin >> K;
  N = 2;
  N += 2;
  is[1][3] = is[3][1] = true;
  is[1][4] = is[4][1] = true;
  wh[0] = 4;
  for (int i = 1; i <= 29; ++i) {
    N += 2;
    is[N][N - 2] = is[N - 2][N] = true;
    is[N][N - 3] = is[N - 3][N] = true;
    is[N - 1][N - 2] = is[N - 2][N - 1] = true;
    is[N - 1][N - 3] = is[N - 3][N - 1] = true;
    wh[i] = N;
  }
  for (int i = 0; (1 << i) <= K; ++i)
    if (K & (1 << i)) {
      ++N;
      is[wh[i]][N] = is[N][wh[i]] = true;
      for (int j = i + 1; j <= 29; ++j) {
        ++N;
        is[N][N - 1] = is[N - 1][N] = true;
      }
      is[N][2] = is[2][N] = true;
    }
  cout << N << '\n';
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j)
      if (is[i][j])
        cout << 'Y';
      else
        cout << 'N';
    cout << '\n';
  }
}

### 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 N, K;
bool is[1002][1002];
int wh[52];
int main() {
  cin >> K;
  N = 2;
  N += 2;
  is[1][3] = is[3][1] = true;
  is[1][4] = is[4][1] = true;
  wh[0] = 4;
  for (int i = 1; i <= 29; ++i) {
    N += 2;
    is[N][N - 2] = is[N - 2][N] = true;
    is[N][N - 3] = is[N - 3][N] = true;
    is[N - 1][N - 2] = is[N - 2][N - 1] = true;
    is[N - 1][N - 3] = is[N - 3][N - 1] = true;
    wh[i] = N;
  }
  for (int i = 0; (1 << i) <= K; ++i)
    if (K & (1 << i)) {
      ++N;
      is[wh[i]][N] = is[N][wh[i]] = true;
      for (int j = i + 1; j <= 29; ++j) {
        ++N;
        is[N][N - 1] = is[N - 1][N] = true;
      }
      is[N][2] = is[2][N] = true;
    }
  cout << N << '\n';
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j)
      if (is[i][j])
        cout << 'Y';
      else
        cout << 'N';
    cout << '\n';
  }
}

```