#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool elem(set<T> st, T val) {
  return (st.find(val) != st.end());
}
template <typename T>
void printv(vector<T> vt) {
  if (vt.size() == 0) return;
  for (int i = 0; i < (int)vt.size() - 1; ++i) cout << vt[i] << " ";
  cout << vt[vt.size() - 1] << endl;
}
vector<string> s;
void add_edge(int a, int b) {
  if (a == b) return;
  s[a][b] = 'Y';
  s[b][a] = 'Y';
}
int arr[31][2];
int main(void) {
  ios::sync_with_stdio(false);
  int k;
  cin >> k;
  string bf = "";
  for (int i = 0; i < (int)100; ++i) bf += "N";
  for (int i = 0; i < (int)100; ++i) {
    s.push_back(bf);
  }
  add_edge(0, 2);
  add_edge(0, 3);
  for (int i = 0; i < (int)30; ++i) {
    int c0 = 2 + i * 2;
    add_edge(c0, c0 + 2);
    add_edge(c0, c0 + 3);
    add_edge(c0 + 1, c0 + 2);
    add_edge(c0 + 1, c0 + 3);
    arr[i + 1][0] = c0;
    arr[i + 1][1] = c0 + 1;
  }
  for (int i = 64; i < 94; ++i) add_edge(i, i + 1);
  add_edge(94, 1);
  for (int i = 0; i < (int)30; ++i) {
    if (i == 0) {
      if (k % 2 == 1) add_edge(0, 64);
      continue;
    }
    if ((k & (1 << i)) > 0) {
      add_edge(arr[i][0], 64 + i);
      add_edge(arr[i][1], 64 + i);
    }
  }
  cout << 100 << endl;
  for (int i = 0; i < (int)100; ++i) cout << s[i] << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename T>
bool elem(set<T> st, T val) {
  return (st.find(val) != st.end());
}
template <typename T>
void printv(vector<T> vt) {
  if (vt.size() == 0) return;
  for (int i = 0; i < (int)vt.size() - 1; ++i) cout << vt[i] << " ";
  cout << vt[vt.size() - 1] << endl;
}
vector<string> s;
void add_edge(int a, int b) {
  if (a == b) return;
  s[a][b] = 'Y';
  s[b][a] = 'Y';
}
int arr[31][2];
int main(void) {
  ios::sync_with_stdio(false);
  int k;
  cin >> k;
  string bf = "";
  for (int i = 0; i < (int)100; ++i) bf += "N";
  for (int i = 0; i < (int)100; ++i) {
    s.push_back(bf);
  }
  add_edge(0, 2);
  add_edge(0, 3);
  for (int i = 0; i < (int)30; ++i) {
    int c0 = 2 + i * 2;
    add_edge(c0, c0 + 2);
    add_edge(c0, c0 + 3);
    add_edge(c0 + 1, c0 + 2);
    add_edge(c0 + 1, c0 + 3);
    arr[i + 1][0] = c0;
    arr[i + 1][1] = c0 + 1;
  }
  for (int i = 64; i < 94; ++i) add_edge(i, i + 1);
  add_edge(94, 1);
  for (int i = 0; i < (int)30; ++i) {
    if (i == 0) {
      if (k % 2 == 1) add_edge(0, 64);
      continue;
    }
    if ((k & (1 << i)) > 0) {
      add_edge(arr[i][0], 64 + i);
      add_edge(arr[i][1], 64 + i);
    }
  }
  cout << 100 << endl;
  for (int i = 0; i < (int)100; ++i) cout << s[i] << endl;
  return 0;
}

```