#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int n, pos[30];
vector<int> vec;
char ans[1001][1001];
int main() {
  cin >> n;
  if (n == 1) {
    printf("2\nNY\nYN");
    return 0;
  }
  for (int i = 0; i < 1001; i++)
    for (int j = 0; j < 1001; j++) ans[i][j] = 'N';
  for (int i = 0; i < 30; i++)
    if (n & (1 << i)) vec.push_back(i);
  int mx = vec[vec.size() - 1];
  pos[0] = 0, pos[mx] = 1;
  int ind = 2, l = 0;
  for (int i = 0; i < mx - 1; i++) {
    ans[l][ind] = ans[ind][l] = 'Y';
    ind++;
    ans[l][ind] = ans[ind][l] = 'Y';
    ind++;
    ans[ind - 1][ind] = ans[ind][ind - 1] = 'Y';
    ans[ind - 2][ind] = ans[ind][ind - 2] = 'Y';
    pos[i + 1] = ind;
    l = ind++;
  }
  ans[l][ind] = ans[ind][l] = 'Y';
  ind++;
  ans[l][ind] = ans[ind][l] = 'Y';
  ind++;
  ans[ind - 1][1] = ans[1][ind - 1] = 'Y';
  ans[ind - 2][1] = ans[1][ind - 2] = 'Y';
  for (int i = 0; i < vec.size() - 1; i++) {
    int x = vec[i];
    int y = mx - x;
    l = 0;
    for (int j = 0; j < y * 2 - 1; j++)
      ans[l][ind] = ans[ind][l] = 'Y', l = ind++;
    ans[l][pos[y]] = ans[pos[y]][l] = 'Y';
  }
  cout << ind << endl;
  for (int i = 0; i < ind; i++) {
    for (int j = 0; j < ind; j++) cout << ans[i][j];
    cout << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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:100000000000000")
using namespace std;
int n, pos[30];
vector<int> vec;
char ans[1001][1001];
int main() {
  cin >> n;
  if (n == 1) {
    printf("2\nNY\nYN");
    return 0;
  }
  for (int i = 0; i < 1001; i++)
    for (int j = 0; j < 1001; j++) ans[i][j] = 'N';
  for (int i = 0; i < 30; i++)
    if (n & (1 << i)) vec.push_back(i);
  int mx = vec[vec.size() - 1];
  pos[0] = 0, pos[mx] = 1;
  int ind = 2, l = 0;
  for (int i = 0; i < mx - 1; i++) {
    ans[l][ind] = ans[ind][l] = 'Y';
    ind++;
    ans[l][ind] = ans[ind][l] = 'Y';
    ind++;
    ans[ind - 1][ind] = ans[ind][ind - 1] = 'Y';
    ans[ind - 2][ind] = ans[ind][ind - 2] = 'Y';
    pos[i + 1] = ind;
    l = ind++;
  }
  ans[l][ind] = ans[ind][l] = 'Y';
  ind++;
  ans[l][ind] = ans[ind][l] = 'Y';
  ind++;
  ans[ind - 1][1] = ans[1][ind - 1] = 'Y';
  ans[ind - 2][1] = ans[1][ind - 2] = 'Y';
  for (int i = 0; i < vec.size() - 1; i++) {
    int x = vec[i];
    int y = mx - x;
    l = 0;
    for (int j = 0; j < y * 2 - 1; j++)
      ans[l][ind] = ans[ind][l] = 'Y', l = ind++;
    ans[l][pos[y]] = ans[pos[y]][l] = 'Y';
  }
  cout << ind << endl;
  for (int i = 0; i < ind; i++) {
    for (int j = 0; j < ind; j++) cout << ans[i][j];
    cout << endl;
  }
  return 0;
}

```