#include <bits/stdc++.h>
using namespace std;
int a[2000][2000], k, p[2000], Index;
void add(int x, int y) { a[x][y] = a[y][x] = 1; }
int main() {
  scanf("%d", &k);
  int maxbit = 0;
  for (int i = (1); i <= (30); ++i)
    if ((1 << i) & k) maxbit = i;
  Index = 2;
  int cur = ++Index;
  add(1, cur);
  for (int i = (1); i <= (maxbit); ++i) {
    add(cur, Index + 1);
    add(cur, Index + 2);
    add(Index + 1, Index + 3);
    add(Index + 2, Index + 3);
    cur = Index += 3;
    p[maxbit - i] = cur;
  }
  add(cur, 2);
  add(1, ++Index);
  cur = Index;
  for (int i = (maxbit - 1); i >= (0); --i) {
    add(cur, ++Index), cur = Index;
    if ((1 << i) & k) add(cur, p[i]);
    add(cur, ++Index), cur = Index;
  }
  printf("%d\n", Index);
  for (int i = (1); i <= (Index); ++i) {
    for (int j = (1); j <= (Index); ++j)
      if (a[i][j])
        printf("Y");
      else
        printf("N");
    puts("");
  }
  assert(Index <= 150);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int a[2000][2000], k, p[2000], Index;
void add(int x, int y) { a[x][y] = a[y][x] = 1; }
int main() {
  scanf("%d", &k);
  int maxbit = 0;
  for (int i = (1); i <= (30); ++i)
    if ((1 << i) & k) maxbit = i;
  Index = 2;
  int cur = ++Index;
  add(1, cur);
  for (int i = (1); i <= (maxbit); ++i) {
    add(cur, Index + 1);
    add(cur, Index + 2);
    add(Index + 1, Index + 3);
    add(Index + 2, Index + 3);
    cur = Index += 3;
    p[maxbit - i] = cur;
  }
  add(cur, 2);
  add(1, ++Index);
  cur = Index;
  for (int i = (maxbit - 1); i >= (0); --i) {
    add(cur, ++Index), cur = Index;
    if ((1 << i) & k) add(cur, p[i]);
    add(cur, ++Index), cur = Index;
  }
  printf("%d\n", Index);
  for (int i = (1); i <= (Index); ++i) {
    for (int j = (1); j <= (Index); ++j)
      if (a[i][j])
        printf("Y");
      else
        printf("N");
    puts("");
  }
  assert(Index <= 150);
  return 0;
}

```