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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long k;
  scanf("%ld", &k);
  long n = 1, m = 0;
  while (n <= k) {
    n *= 2;
    m++;
  }
  m--;
  long d[1000][1000];
  for (int i = 0; i < 500; i++)
    for (int j = 0; j < 500; j++) d[i][j] = 0;
  long current = 0;
  long next = 2;
  long depth = 0;
  while (k) {
    if (k == 1) {
      d[current][1] = 1;
      d[1][current] = 1;
      break;
    }
    if (k % 2) {
      d[current][next] = 1;
      d[next][current] = 1;
      for (int i = 0; i < 2 * m - depth - 1; i++) {
        d[next][next + 1] = 1;
        d[next + 1][next] = 1;
        next++;
      }
      d[next][1] = 1;
      d[1][next] = 1;
      next++;
    }
    d[current][next] = 1;
    d[next][current] = 1;
    d[next][next + 2] = 1;
    d[next + 2][next] = 1;
    d[current][next + 1] = 1;
    d[next + 1][current] = 1;
    d[next + 1][next + 2] = 1;
    d[next + 2][next + 1] = 1;
    current = next + 2;
    next = next + 3;
    k /= 2;
    depth += 2;
  }
  cout << next << endl;
  for (int i = 0; i < next; i++) {
    for (int j = 0; j < next; j++)
      if (d[i][j]) {
        printf("Y");
      } else
        printf("N");
    printf("\n");
  }
  return 0;
}
