#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
vector<int> vec[1000];
char s[1010][1010];
void addedge(int a, int b) {
  vec[a].push_back(b);
  vec[b].push_back(a);
}
int main() {
  int k;
  cin >> k;
  int x = 0, y = 0, z = 2;
  while (k) {
    if (k & 1) {
      addedge(x, z);
      addedge(y, z);
    }
    k >>= 1;
    if (k == 0) break;
    addedge(x, z + 1);
    addedge(x, z + 2);
    addedge(y, z + 1);
    addedge(y, z + 2);
    addedge(z, z + 3);
    x = z + 1;
    y = z + 2;
    z = z + 3;
  }
  addedge(1, z);
  memset(s, 'N', sizeof s);
  cout << z + 1 << endl;
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j < vec[i].size(); j++) {
      s[i][vec[i][j]] = 'Y';
      s[vec[i][j]][i] = 'Y';
    }
  }
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j <= z; j++) {
      printf("%c", s[i][j]);
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
vector<int> vec[1000];
char s[1010][1010];
void addedge(int a, int b) {
  vec[a].push_back(b);
  vec[b].push_back(a);
}
int main() {
  int k;
  cin >> k;
  int x = 0, y = 0, z = 2;
  while (k) {
    if (k & 1) {
      addedge(x, z);
      addedge(y, z);
    }
    k >>= 1;
    if (k == 0) break;
    addedge(x, z + 1);
    addedge(x, z + 2);
    addedge(y, z + 1);
    addedge(y, z + 2);
    addedge(z, z + 3);
    x = z + 1;
    y = z + 2;
    z = z + 3;
  }
  addedge(1, z);
  memset(s, 'N', sizeof s);
  cout << z + 1 << endl;
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j < vec[i].size(); j++) {
      s[i][vec[i][j]] = 'Y';
      s[vec[i][j]][i] = 'Y';
    }
  }
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j <= z; j++) {
      printf("%c", s[i][j]);
    }
    cout << endl;
  }
  return 0;
}

```