#include <bits/stdc++.h>
using std::cin;
using std::cout;
using std::endl;
using std::list;
using std::vector;
using std::rel_ops::operator>;
using namespace std;
const int INF = int(1e9);
const long long INF64 = (long long)(INF) * (long long)(INF);
const long double EPS = 1e-9;
const long double PI = (long double)(3.1415926535897932384626433832795);
int g[1002][1002];
void make(int a, int b) { g[a][b] = g[b][a] = 1; }
int main() {
  int k;
  cin >> k;
  for (int i = 3; i < 80; ++i) {
    if (i % 2) {
      make(i, i + 2);
      make(i, i + 3);
    } else {
      make(i, i + 1);
      make(i, i + 2);
    }
  }
  make(1, 3);
  make(1, 4);
  for (int i = 100; i < 150; ++i) {
    make(i, i + 1);
  }
  make(150, 2);
  for (int i = 0; i < 32; ++i) {
    if (k & (1 << i)) {
      make(4 + 2 * i, 100 + i);
    }
  }
  cout << 150 << endl;
  for (int i = 1; i <= 150; ++i) {
    for (int j = 1; j <= 150; ++j) {
      cout << (g[i][j] ? 'Y' : 'N');
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 std::cin;
using std::cout;
using std::endl;
using std::list;
using std::vector;
using std::rel_ops::operator>;
using namespace std;
const int INF = int(1e9);
const long long INF64 = (long long)(INF) * (long long)(INF);
const long double EPS = 1e-9;
const long double PI = (long double)(3.1415926535897932384626433832795);
int g[1002][1002];
void make(int a, int b) { g[a][b] = g[b][a] = 1; }
int main() {
  int k;
  cin >> k;
  for (int i = 3; i < 80; ++i) {
    if (i % 2) {
      make(i, i + 2);
      make(i, i + 3);
    } else {
      make(i, i + 1);
      make(i, i + 2);
    }
  }
  make(1, 3);
  make(1, 4);
  for (int i = 100; i < 150; ++i) {
    make(i, i + 1);
  }
  make(150, 2);
  for (int i = 0; i < 32; ++i) {
    if (k & (1 << i)) {
      make(4 + 2 * i, 100 + i);
    }
  }
  cout << 150 << endl;
  for (int i = 1; i <= 150; ++i) {
    for (int j = 1; j <= 150; ++j) {
      cout << (g[i][j] ? 'Y' : 'N');
    }
    cout << endl;
  }
  return 0;
}

```