#include <bits/stdc++.h>
using namespace std;
bool A[1000][1000];
vector<short> B;
int main() {
  int k;
  short a = 0, b = 0, c = 2;
  for (scanf("%d", &k); k > 1; k >>= 1) {
    for (vector<short>::iterator j = B.begin(); j != B.end(); j++) {
      A[*j][c] = true;
      *j = c;
      c++;
    }
    if (k & 1) {
      A[a][c] = A[b][c] = true;
      A[a][c + 1] = A[b][c + 1] = true;
      A[a][c + 2] = A[b][c + 2] = true;
      B.push_back(c);
      a = c + 1;
      b = c + 2;
      c += 3;
    } else {
      A[a][c] = A[b][c] = true;
      A[a][c + 1] = A[b][c + 1] = true;
      a = c;
      b = c + 1;
      c += 2;
    }
  }
  for (vector<short>::iterator j = B.begin(); j != B.end(); j++)
    A[*j][1] = true;
  A[a][1] = A[b][1] = true;
  printf("%hd\n", c);
  for (int i = 0; i < c; i++) {
    for (int j = 0; j < c; j++) putchar(A[i][j] || A[j][i] ? 'Y' : 'N');
    putchar('\n');
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
bool A[1000][1000];
vector<short> B;
int main() {
  int k;
  short a = 0, b = 0, c = 2;
  for (scanf("%d", &k); k > 1; k >>= 1) {
    for (vector<short>::iterator j = B.begin(); j != B.end(); j++) {
      A[*j][c] = true;
      *j = c;
      c++;
    }
    if (k & 1) {
      A[a][c] = A[b][c] = true;
      A[a][c + 1] = A[b][c + 1] = true;
      A[a][c + 2] = A[b][c + 2] = true;
      B.push_back(c);
      a = c + 1;
      b = c + 2;
      c += 3;
    } else {
      A[a][c] = A[b][c] = true;
      A[a][c + 1] = A[b][c + 1] = true;
      a = c;
      b = c + 1;
      c += 2;
    }
  }
  for (vector<short>::iterator j = B.begin(); j != B.end(); j++)
    A[*j][1] = true;
  A[a][1] = A[b][1] = true;
  printf("%hd\n", c);
  for (int i = 0; i < c; i++) {
    for (int j = 0; j < c; j++) putchar(A[i][j] || A[j][i] ? 'Y' : 'N');
    putchar('\n');
  }
  return 0;
}

```