#include <bits/stdc++.h>
const long long inf = 9000000000000000000ll;
using namespace std;
int n, k, b[3000];
bool s[3000][3000];
pair<int, int> a[3000];
int main() {
  cin >> k;
  memset(s, false, sizeof(s));
  s[1][3] = true;
  s[3][1] = true;
  s[1][4] = true;
  s[4][1] = true;
  a[1] = make_pair(3, 4);
  n = 4;
  for (int i = 1; i <= 29; i++) {
    n++;
    s[n][n - 1] = true;
    s[n - 1][n] = true;
    s[n][n - 2] = true;
    s[n - 2][n] = true;
    n++;
    s[n][n - 2] = true;
    s[n - 2][n] = true;
    s[n][n - 3] = true;
    s[n - 3][n] = true;
    a[i + 1] = make_pair(n - 1, n);
  }
  n++;
  b[0] = n;
  for (int i = 1; i <= 29; i++) {
    n++;
    s[n][n - 1] = true;
    s[n - 1][n] = true;
    b[i] = n;
  }
  s[2][n] = true;
  s[n][2] = true;
  for (int i = 30; i >= 0; i--)
    if ((k | (1 << i)) == k) {
      if (i != 0) {
        int v1 = a[i].first, v2 = a[i].second, v3 = b[i];
        s[v1][v3] = true;
        s[v3][v1] = true;
        s[v2][v3] = true;
        s[v3][v2] = true;
      } else {
        s[1][b[0]] = true;
        s[b[0]][1] = true;
      }
    }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if (s[i][j])
        printf("%c", 'Y');
      else
        printf("%c", 'N');
    printf("\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>
const long long inf = 9000000000000000000ll;
using namespace std;
int n, k, b[3000];
bool s[3000][3000];
pair<int, int> a[3000];
int main() {
  cin >> k;
  memset(s, false, sizeof(s));
  s[1][3] = true;
  s[3][1] = true;
  s[1][4] = true;
  s[4][1] = true;
  a[1] = make_pair(3, 4);
  n = 4;
  for (int i = 1; i <= 29; i++) {
    n++;
    s[n][n - 1] = true;
    s[n - 1][n] = true;
    s[n][n - 2] = true;
    s[n - 2][n] = true;
    n++;
    s[n][n - 2] = true;
    s[n - 2][n] = true;
    s[n][n - 3] = true;
    s[n - 3][n] = true;
    a[i + 1] = make_pair(n - 1, n);
  }
  n++;
  b[0] = n;
  for (int i = 1; i <= 29; i++) {
    n++;
    s[n][n - 1] = true;
    s[n - 1][n] = true;
    b[i] = n;
  }
  s[2][n] = true;
  s[n][2] = true;
  for (int i = 30; i >= 0; i--)
    if ((k | (1 << i)) == k) {
      if (i != 0) {
        int v1 = a[i].first, v2 = a[i].second, v3 = b[i];
        s[v1][v3] = true;
        s[v3][v1] = true;
        s[v2][v3] = true;
        s[v3][v2] = true;
      } else {
        s[1][b[0]] = true;
        s[b[0]][1] = true;
      }
    }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if (s[i][j])
        printf("%c", 'Y');
      else
        printf("%c", 'N');
    printf("\n");
  }
  return 0;
}

```