#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1000, Length = 62;
int Who[Length], K, Current;
char v[Maxn][Maxn];
void Create(int Fi, int Se) { v[Fi][Se] = v[Se][Fi] = 'Y'; }
int main() {
  memset(v, 'N', sizeof v);
  Current = 3;
  Create(1, Current);
  while (Current++ < Length) Create(Current, Current - 1);
  Current--;
  Create(2, Current + 1);
  Create(2, Current + 2);
  Create(Current + 1, Current + 3);
  Create(Current + 2, Current + 3);
  Current += 3;
  Who[0] = 2;
  Who[1] = Current;
  for (int i = 2; i <= Length / 2; i++) {
    Create(Current, Current + 1);
    Create(Current, Current + 2);
    Create(Current + 1, Current + 3);
    Create(Current + 2, Current + 3);
    Current += 3;
    Who[i] = Current;
  }
  scanf("%d", &K);
  int Now = Length;
  for (int i = 0; i < 30; i++, Now -= 2)
    if (K & (1 << i)) Create(Now, Who[i]);
  printf("%d\n", Current);
  for (int i = 1; i <= Current; i++) {
    for (int j = 1; j <= Current; j++) putchar(v[i][j]);
    printf("\n");
  }
}

### Prompt

Generate 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;
const int Maxn = 1000, Length = 62;
int Who[Length], K, Current;
char v[Maxn][Maxn];
void Create(int Fi, int Se) { v[Fi][Se] = v[Se][Fi] = 'Y'; }
int main() {
  memset(v, 'N', sizeof v);
  Current = 3;
  Create(1, Current);
  while (Current++ < Length) Create(Current, Current - 1);
  Current--;
  Create(2, Current + 1);
  Create(2, Current + 2);
  Create(Current + 1, Current + 3);
  Create(Current + 2, Current + 3);
  Current += 3;
  Who[0] = 2;
  Who[1] = Current;
  for (int i = 2; i <= Length / 2; i++) {
    Create(Current, Current + 1);
    Create(Current, Current + 2);
    Create(Current + 1, Current + 3);
    Create(Current + 2, Current + 3);
    Current += 3;
    Who[i] = Current;
  }
  scanf("%d", &K);
  int Now = Length;
  for (int i = 0; i < 30; i++, Now -= 2)
    if (K & (1 << i)) Create(Now, Who[i]);
  printf("%d\n", Current);
  for (int i = 1; i <= Current; i++) {
    for (int j = 1; j <= Current; j++) putchar(v[i][j]);
    printf("\n");
  }
}

```