#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
const int N = 1005;
bool adj[N][N];
inline void add(int u, int v) { adj[u][v] = adj[v][u] = true; }
int main() {
  int k;
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY\nYN\n";
    return 0;
  }
  vector<PII> inter;
  add(2, 3);
  int cur = 4;
  vector<VI> layers{{1}};
  vector<bool> jmp{false};
  while (k > 1) {
    jmp.push_back(k & 1);
    layers.push_back({cur, cur + 1});
    cur += 2;
    k /= 2;
  }
  layers.push_back({3});
  jmp.push_back(false);
  int nn = layers.size();
  for (int i = 0; i + 1 < nn; i++) {
    for (int u : layers[i]) {
      for (int v : layers[i + 1]) {
        add(u, v);
      }
    }
  }
  for (int i = 1; i < nn - 1; i++) {
    int s = layers[i][0];
    if (jmp[i]) {
      int len = nn - i;
      while (len > 1) {
        len--;
        add(s, cur);
        s = cur;
        cur++;
      }
      add(s, 2);
    }
  }
  const int n = cur - 1;
  cout << n << endl;
  for (int u = 1; u <= n; u++) {
    for (int v = 1; v <= n; ++v) {
      cout << (adj[u][v] ? 'Y' : 'N');
    }
    cout << '\n';
  }
  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;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
const int N = 1005;
bool adj[N][N];
inline void add(int u, int v) { adj[u][v] = adj[v][u] = true; }
int main() {
  int k;
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY\nYN\n";
    return 0;
  }
  vector<PII> inter;
  add(2, 3);
  int cur = 4;
  vector<VI> layers{{1}};
  vector<bool> jmp{false};
  while (k > 1) {
    jmp.push_back(k & 1);
    layers.push_back({cur, cur + 1});
    cur += 2;
    k /= 2;
  }
  layers.push_back({3});
  jmp.push_back(false);
  int nn = layers.size();
  for (int i = 0; i + 1 < nn; i++) {
    for (int u : layers[i]) {
      for (int v : layers[i + 1]) {
        add(u, v);
      }
    }
  }
  for (int i = 1; i < nn - 1; i++) {
    int s = layers[i][0];
    if (jmp[i]) {
      int len = nn - i;
      while (len > 1) {
        len--;
        add(s, cur);
        s = cur;
        cur++;
      }
      add(s, 2);
    }
  }
  const int n = cur - 1;
  cout << n << endl;
  for (int u = 1; u <= n; u++) {
    for (int v = 1; v <= n; ++v) {
      cout << (adj[u][v] ? 'Y' : 'N');
    }
    cout << '\n';
  }
  return 0;
}

```