#include <bits/stdc++.h>
using namespace std;
int_fast64_t n, p, x, i, v[1002][1002], v2[1002], last1, last2, nr, m, j;
int main() {
  cin >> n;
  if (n == 1) {
    cout << 2 << '\n';
    cout << "NY" << '\n' << "YN";
    return 0;
  }
  p = 1;
  x = 0;
  while (p <= n) {
    p *= 2;
    x++;
  }
  x--;
  p /= 2;
  v[1][3] = 1;
  v[3][1] = 1;
  v[1][4] = 1;
  v[4][1] = 1;
  v2[x] = 3;
  last1 = 3;
  last2 = 4;
  nr = 5;
  for (i = 2; i <= x; i++) {
    v[last1][nr] = 1;
    v[last1][nr + 1] = 1;
    v[nr][last1] = 1;
    v[nr + 1][last1] = 1;
    v[last2][nr] = 1;
    v[last2][nr + 1] = 1;
    v[nr][last2] = 1;
    v[nr + 1][last2] = 1;
    v2[x - i + 1] = nr;
    last1 = nr;
    last2 = nr + 1;
    nr += 2;
  }
  v[last1][2] = 1;
  v[last2][2] = 1;
  v[2][last1] = 1;
  v[2][last2] = 1;
  n -= p;
  while (n) {
    if (n < p) {
      p /= 2;
      x--;
      m++;
    } else if (x != 0) {
      n -= p;
      v[1][nr] = 1;
      v[nr][1] = 1;
      for (i = 2; i <= m; i++) {
        v[nr][nr + 1] = 1;
        v[nr + 1][nr] = 1;
        nr++;
      }
      v[nr][v2[x]] = 1;
      v[nr][v2[x] + 1] = 1;
      v[v2[x]][nr] = 1;
      v[v2[x] + 1][nr] = 1;
      nr++;
    } else {
      n -= p;
      p /= 2;
      x--;
      v[1][nr] = 1;
      v[nr][1] = 1;
      for (i = 2; i <= m; i++) {
        v[nr][nr + 1] = 1;
        v[nr + 1][nr] = 1;
        nr++;
      }
      v[nr][2] = 1;
      v[2][nr] = 1;
    }
  }
  cout << nr << '\n';
  for (i = 1; i <= nr; i++) {
    for (j = 1; j <= nr; j++) {
      if (v[i][j] == 1) {
        cout << "Y";
      } else {
        cout << "N";
      }
    }
    cout << '\n';
  }
  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;
int_fast64_t n, p, x, i, v[1002][1002], v2[1002], last1, last2, nr, m, j;
int main() {
  cin >> n;
  if (n == 1) {
    cout << 2 << '\n';
    cout << "NY" << '\n' << "YN";
    return 0;
  }
  p = 1;
  x = 0;
  while (p <= n) {
    p *= 2;
    x++;
  }
  x--;
  p /= 2;
  v[1][3] = 1;
  v[3][1] = 1;
  v[1][4] = 1;
  v[4][1] = 1;
  v2[x] = 3;
  last1 = 3;
  last2 = 4;
  nr = 5;
  for (i = 2; i <= x; i++) {
    v[last1][nr] = 1;
    v[last1][nr + 1] = 1;
    v[nr][last1] = 1;
    v[nr + 1][last1] = 1;
    v[last2][nr] = 1;
    v[last2][nr + 1] = 1;
    v[nr][last2] = 1;
    v[nr + 1][last2] = 1;
    v2[x - i + 1] = nr;
    last1 = nr;
    last2 = nr + 1;
    nr += 2;
  }
  v[last1][2] = 1;
  v[last2][2] = 1;
  v[2][last1] = 1;
  v[2][last2] = 1;
  n -= p;
  while (n) {
    if (n < p) {
      p /= 2;
      x--;
      m++;
    } else if (x != 0) {
      n -= p;
      v[1][nr] = 1;
      v[nr][1] = 1;
      for (i = 2; i <= m; i++) {
        v[nr][nr + 1] = 1;
        v[nr + 1][nr] = 1;
        nr++;
      }
      v[nr][v2[x]] = 1;
      v[nr][v2[x] + 1] = 1;
      v[v2[x]][nr] = 1;
      v[v2[x] + 1][nr] = 1;
      nr++;
    } else {
      n -= p;
      p /= 2;
      x--;
      v[1][nr] = 1;
      v[nr][1] = 1;
      for (i = 2; i <= m; i++) {
        v[nr][nr + 1] = 1;
        v[nr + 1][nr] = 1;
        nr++;
      }
      v[nr][2] = 1;
      v[2][nr] = 1;
    }
  }
  cout << nr << '\n';
  for (i = 1; i <= nr; i++) {
    for (j = 1; j <= nr; j++) {
      if (v[i][j] == 1) {
        cout << "Y";
      } else {
        cout << "N";
      }
    }
    cout << '\n';
  }
  return 0;
}

```