#include <bits/stdc++.h>
using namespace std;
int n = 1000, k;
bool a[1000][1000];
int s = 0, f = 1;
int cnt = 2;
void add(int x, int y) { a[x][y] = a[y][x] = true; }
void make(int k) {
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(s, i);
  for (int(ttt) = (0); ttt < (4); ++(ttt)) {
    for (int(i) = (cnt); i < (cnt + k); ++(i))
      for (int(j) = (cnt + k); j < (cnt + k + k); ++(j)) add(i, j);
    cnt += k;
  }
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(i, f);
  cnt += k;
}
void m2(int k) {
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(s, i);
  for (int(ttt) = (0); ttt < (2); ++(ttt)) {
    for (int(i) = (cnt); i < (cnt + k); ++(i))
      for (int(j) = (cnt + k); j < (cnt + k + k); ++(j)) add(i, j);
    cnt += k;
  }
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(i, cnt + k);
  add(cnt + k, cnt + k + 1);
  add(cnt + k + 1, f);
  cnt += k + 2;
}
int main() {
  cin >> k;
  while (k) {
    int r = (int)pow(k, 1.0 / 5);
    if (r > 1) {
      make(r);
      k -= r * r * r * r * r;
    } else {
      r = pow(k, 1.0 / 3);
      m2(r);
      k -= r * r * r;
    }
  }
  n = cnt;
  cout << n << "\n";
  for (int(i) = (0); i < (n); ++(i)) {
    for (int(j) = (0); j < (n); ++(j)) cout << (a[i][j] == true ? 'Y' : 'N');
    cout << "\n";
  }
  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>
using namespace std;
int n = 1000, k;
bool a[1000][1000];
int s = 0, f = 1;
int cnt = 2;
void add(int x, int y) { a[x][y] = a[y][x] = true; }
void make(int k) {
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(s, i);
  for (int(ttt) = (0); ttt < (4); ++(ttt)) {
    for (int(i) = (cnt); i < (cnt + k); ++(i))
      for (int(j) = (cnt + k); j < (cnt + k + k); ++(j)) add(i, j);
    cnt += k;
  }
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(i, f);
  cnt += k;
}
void m2(int k) {
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(s, i);
  for (int(ttt) = (0); ttt < (2); ++(ttt)) {
    for (int(i) = (cnt); i < (cnt + k); ++(i))
      for (int(j) = (cnt + k); j < (cnt + k + k); ++(j)) add(i, j);
    cnt += k;
  }
  for (int(i) = (cnt); i < (cnt + k); ++(i)) add(i, cnt + k);
  add(cnt + k, cnt + k + 1);
  add(cnt + k + 1, f);
  cnt += k + 2;
}
int main() {
  cin >> k;
  while (k) {
    int r = (int)pow(k, 1.0 / 5);
    if (r > 1) {
      make(r);
      k -= r * r * r * r * r;
    } else {
      r = pow(k, 1.0 / 3);
      m2(r);
      k -= r * r * r;
    }
  }
  n = cnt;
  cout << n << "\n";
  for (int(i) = (0); i < (n); ++(i)) {
    for (int(j) = (0); j < (n); ++(j)) cout << (a[i][j] == true ? 'Y' : 'N');
    cout << "\n";
  }
  return 0;
}

```