#include <bits/stdc++.h>
using namespace std;
const int max_n = 1000;
bool G[max_n][max_n];
vector<int> bit;
void edge(int u, int v) { G[u][v] = G[v][u] = true; }
int main() {
  int k;
  cin >> k;
  for (; k > 0; k = k >> 1) bit.push_back(k & 1);
  const int lg = bit.size() - 1;
  memset(G, 0, sizeof G);
  if (bit[0]) edge(2, 0);
  int n = 3;
  for (int i = 1; i <= lg; i++) {
    edge(n - 1, n);
    n++;
  }
  edge(n - 1, 1);
  for (int i = 1; i <= lg; i++)
    if (bit[i])
      for (int j = 1; j <= i; j++) {
        if (j == 1)
          edge(n, 0), edge(n + 1, 0);
        else {
          edge(n, n - 1), edge(n + 1, n - 1);
          edge(n, n - 2), edge(n + 1, n - 2);
        }
        if (i == j) edge(n, i + 2), edge(n + 1, i + 2);
        n += 2;
      }
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << (G[i][j] ? 'Y' : 'N');
    cout << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int max_n = 1000;
bool G[max_n][max_n];
vector<int> bit;
void edge(int u, int v) { G[u][v] = G[v][u] = true; }
int main() {
  int k;
  cin >> k;
  for (; k > 0; k = k >> 1) bit.push_back(k & 1);
  const int lg = bit.size() - 1;
  memset(G, 0, sizeof G);
  if (bit[0]) edge(2, 0);
  int n = 3;
  for (int i = 1; i <= lg; i++) {
    edge(n - 1, n);
    n++;
  }
  edge(n - 1, 1);
  for (int i = 1; i <= lg; i++)
    if (bit[i])
      for (int j = 1; j <= i; j++) {
        if (j == 1)
          edge(n, 0), edge(n + 1, 0);
        else {
          edge(n, n - 1), edge(n + 1, n - 1);
          edge(n, n - 2), edge(n + 1, n - 2);
        }
        if (i == j) edge(n, i + 2), edge(n + 1, i + 2);
        n += 2;
      }
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << (G[i][j] ? 'Y' : 'N');
    cout << endl;
  }
  return 0;
}

```