#include <bits/stdc++.h>
using namespace std;
char gr[1000][1000 + 1];
int main() {
  memset(gr, 'N', sizeof gr);
  int k;
  scanf("%d", &k);
  int acc = 0, fv = 3, pt = 1;
  while (acc != k) {
    int ptv = fv + (acc > 0);
    int nfv = ptv + 1;
    if (acc) gr[fv][nfv] = gr[nfv][fv] = 'Y';
    int nptv = nfv + (acc > 0);
    gr[ptv][nptv] = gr[nptv][ptv] = 'Y';
    gr[ptv][nptv + 1] = gr[nptv + 1][ptv] = 'Y';
    int nnfv = nfv + (acc > 0) + 2;
    if (acc) gr[nfv][nnfv] = gr[nnfv][nfv] = 'Y';
    if (k & pt) {
      gr[nptv][nnfv] = gr[nnfv][nptv] = 'Y';
      acc += pt;
    }
    int nnptv = nnfv + (acc > 0);
    gr[nptv][nnptv] = gr[nnptv][nptv] = 'Y';
    gr[nptv + 1][nnptv] = gr[nnptv][nptv + 1] = 'Y';
    fv = nnfv;
    pt <<= 1;
  }
  gr[1][3] = gr[3][1] = 'Y';
  gr[fv][2] = gr[2][fv] = 'Y';
  int n = fv + 1 - 2 + 2;
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      printf("%c", gr[i][j]);
    }
    puts("");
  }
  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>
using namespace std;
char gr[1000][1000 + 1];
int main() {
  memset(gr, 'N', sizeof gr);
  int k;
  scanf("%d", &k);
  int acc = 0, fv = 3, pt = 1;
  while (acc != k) {
    int ptv = fv + (acc > 0);
    int nfv = ptv + 1;
    if (acc) gr[fv][nfv] = gr[nfv][fv] = 'Y';
    int nptv = nfv + (acc > 0);
    gr[ptv][nptv] = gr[nptv][ptv] = 'Y';
    gr[ptv][nptv + 1] = gr[nptv + 1][ptv] = 'Y';
    int nnfv = nfv + (acc > 0) + 2;
    if (acc) gr[nfv][nnfv] = gr[nnfv][nfv] = 'Y';
    if (k & pt) {
      gr[nptv][nnfv] = gr[nnfv][nptv] = 'Y';
      acc += pt;
    }
    int nnptv = nnfv + (acc > 0);
    gr[nptv][nnptv] = gr[nnptv][nptv] = 'Y';
    gr[nptv + 1][nnptv] = gr[nnptv][nptv + 1] = 'Y';
    fv = nnfv;
    pt <<= 1;
  }
  gr[1][3] = gr[3][1] = 'Y';
  gr[fv][2] = gr[2][fv] = 'Y';
  int n = fv + 1 - 2 + 2;
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      printf("%c", gr[i][j]);
    }
    puts("");
  }
  return 0;
}

```