#include <bits/stdc++.h>
using namespace std;
long long G(long long a, long long b) {
  if (b == 0) return a;
  return G(b, a % b);
}
vector<long long> v[40];
bool mark[1111][1111];
int main() {
  ios_base::sync_with_stdio(false);
  long long k;
  cin >> k;
  string s = "";
  long long t = k;
  while (t) {
    if (t % 2)
      s += "1";
    else
      s += "0";
    t /= 2;
  }
  long long n = 2, d = 1;
  v[0].push_back(0);
  reverse(s.begin(), s.end());
  for (int i = 0; i < s.length(); i++) {
    if (i) {
      v[d].push_back(n);
      v[d].push_back(n + 1);
      for (int j = 0; j < v[d - 1].size(); j++) {
        mark[n][v[d - 1][j]] = mark[v[d - 1][j]][n] = 1;
        mark[n + 1][v[d - 1][j]] = mark[v[d - 1][j]][n + 1] = 1;
      }
      n += 2;
      d++;
    }
    if (s[i] == '1') {
      for (int j = 1; j < d; j++) {
        v[j].push_back(n);
        if (j > 1)
          mark[n][n - 1] = mark[n - 1][n] = 1;
        else
          mark[n][0] = mark[0][n] = 1;
        n++;
      }
    }
  }
  for (int i = 0; i < v[d - 1].size(); i++)
    mark[1][v[d - 1][i]] = mark[v[d - 1][i]][1] = 1;
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << (mark[i][j] ? "Y" : "N");
    }
    cout << endl;
  }
}

### 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;
long long G(long long a, long long b) {
  if (b == 0) return a;
  return G(b, a % b);
}
vector<long long> v[40];
bool mark[1111][1111];
int main() {
  ios_base::sync_with_stdio(false);
  long long k;
  cin >> k;
  string s = "";
  long long t = k;
  while (t) {
    if (t % 2)
      s += "1";
    else
      s += "0";
    t /= 2;
  }
  long long n = 2, d = 1;
  v[0].push_back(0);
  reverse(s.begin(), s.end());
  for (int i = 0; i < s.length(); i++) {
    if (i) {
      v[d].push_back(n);
      v[d].push_back(n + 1);
      for (int j = 0; j < v[d - 1].size(); j++) {
        mark[n][v[d - 1][j]] = mark[v[d - 1][j]][n] = 1;
        mark[n + 1][v[d - 1][j]] = mark[v[d - 1][j]][n + 1] = 1;
      }
      n += 2;
      d++;
    }
    if (s[i] == '1') {
      for (int j = 1; j < d; j++) {
        v[j].push_back(n);
        if (j > 1)
          mark[n][n - 1] = mark[n - 1][n] = 1;
        else
          mark[n][0] = mark[0][n] = 1;
        n++;
      }
    }
  }
  for (int i = 0; i < v[d - 1].size(); i++)
    mark[1][v[d - 1][i]] = mark[v[d - 1][i]][1] = 1;
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << (mark[i][j] ? "Y" : "N");
    }
    cout << endl;
  }
}

```