#include <bits/stdc++.h>
const long long MOD = 1e9 + 7;
using pii = std::pair<long long, long long>;
using namespace std;
long long k;
long long exists[1005][1005] = {0};
class BFoxAndMinimalPath {
 public:
  void solve(std::istream& in, std::ostream& out) {
    in >> k;
    long long cur = 3, length = 63 - __builtin_clzll(k);
    for (long long i = (0); i < (length); i++) {
      exists[cur + 3 * i][cur + 3 * i + 1] = 1;
      exists[cur + 3 * i][cur + 3 * i + 2] = 1;
      exists[cur + 3 * (i + 1)][cur + 3 * i + 1] = 1;
      exists[cur + 3 * (i + 1)][cur + 3 * i + 2] = 1;
    }
    exists[2][3 * (length + 1)] = 1;
    cur = 3 * (length + 1) + 1;
    for (long long i = (0); i < (length + 1); i++)
      if (k & (1 << i)) {
        exists[1][cur] = 1;
        for (long long j = (0); j < (2 * (length - i)); j++)
          exists[cur + j][cur + j + 1] = 1;
        exists[cur + 2 * (length - i)][3 + 3 * (length - i)] = 1;
        cur += (2 * (length - i) + 1);
      }
    out << cur - 1 << "\n";
    for (long long i = (1); i < (cur); i++) {
      for (long long j = (1); j < (cur); j++)
        out << "NY"[exists[i][j] || exists[j][i]];
      out << "\n";
    }
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  BFoxAndMinimalPath solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  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>
const long long MOD = 1e9 + 7;
using pii = std::pair<long long, long long>;
using namespace std;
long long k;
long long exists[1005][1005] = {0};
class BFoxAndMinimalPath {
 public:
  void solve(std::istream& in, std::ostream& out) {
    in >> k;
    long long cur = 3, length = 63 - __builtin_clzll(k);
    for (long long i = (0); i < (length); i++) {
      exists[cur + 3 * i][cur + 3 * i + 1] = 1;
      exists[cur + 3 * i][cur + 3 * i + 2] = 1;
      exists[cur + 3 * (i + 1)][cur + 3 * i + 1] = 1;
      exists[cur + 3 * (i + 1)][cur + 3 * i + 2] = 1;
    }
    exists[2][3 * (length + 1)] = 1;
    cur = 3 * (length + 1) + 1;
    for (long long i = (0); i < (length + 1); i++)
      if (k & (1 << i)) {
        exists[1][cur] = 1;
        for (long long j = (0); j < (2 * (length - i)); j++)
          exists[cur + j][cur + j + 1] = 1;
        exists[cur + 2 * (length - i)][3 + 3 * (length - i)] = 1;
        cur += (2 * (length - i) + 1);
      }
    out << cur - 1 << "\n";
    for (long long i = (1); i < (cur); i++) {
      for (long long j = (1); j < (cur); j++)
        out << "NY"[exists[i][j] || exists[j][i]];
      out << "\n";
    }
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  BFoxAndMinimalPath solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```