#include <bits/stdc++.h>
using namespace std;
const int N = 1100;
bool a[N][N];
int main() {
  long long k;
  while (cin >> k) {
    int n = 0;
    while ((1 << n) <= k) n++;
    if (n) n--;
    int step = n * 2 + 1;
    int num = (1 << n);
    memset(a, 0, sizeof(a));
    n = 3 * n + 3;
    a[1][3] = 1;
    for (int i = 4; i <= n; i++) {
      if (i % 3 == 0)
        a[i][i - 1] = a[i][i - 2] = 1;
      else if ((i - 1) % 3 == 0)
        a[i][i - 1] = 1;
      else if ((i - 2) % 3 == 0)
        a[i][i - 2] = 1;
    }
    a[n][2] = 1;
    for (int i = n + 1; i < n + step; i++) a[i][i + 1] = 1;
    a[n + step - 1][2] = 1;
    int tmp = k - num, ptr = 1;
    while (tmp) {
      int fg = (1 & tmp);
      tmp >>= 1;
      if (fg) a[(ptr + 1) / 2 * 3][ptr + n] = 1;
      ptr += 2;
    }
    if (k != num) n += (step - 1);
    cout << n << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++)
        if (a[i][j] || a[j][i])
          cout << 'Y';
        else
          cout << 'N';
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1100;
bool a[N][N];
int main() {
  long long k;
  while (cin >> k) {
    int n = 0;
    while ((1 << n) <= k) n++;
    if (n) n--;
    int step = n * 2 + 1;
    int num = (1 << n);
    memset(a, 0, sizeof(a));
    n = 3 * n + 3;
    a[1][3] = 1;
    for (int i = 4; i <= n; i++) {
      if (i % 3 == 0)
        a[i][i - 1] = a[i][i - 2] = 1;
      else if ((i - 1) % 3 == 0)
        a[i][i - 1] = 1;
      else if ((i - 2) % 3 == 0)
        a[i][i - 2] = 1;
    }
    a[n][2] = 1;
    for (int i = n + 1; i < n + step; i++) a[i][i + 1] = 1;
    a[n + step - 1][2] = 1;
    int tmp = k - num, ptr = 1;
    while (tmp) {
      int fg = (1 & tmp);
      tmp >>= 1;
      if (fg) a[(ptr + 1) / 2 * 3][ptr + n] = 1;
      ptr += 2;
    }
    if (k != num) n += (step - 1);
    cout << n << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++)
        if (a[i][j] || a[j][i])
          cout << 'Y';
        else
          cout << 'N';
      cout << endl;
    }
  }
  return 0;
}

```