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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool a[1001][1001] = {0};
int main() {
  int k = 1, nh = 1, nr = 3, ans = 1, j, i, n, dpth;
  double spl;
  scanf("%d", &n);
  if (n == 1) {
    printf("2\nNY\nYN");
    return 0;
  }
  spl = (double)n;
  dpth = log(spl) / log(2.0);
  if ((n & 1) == 0) {
    for (i = nr; i < nr + 2; i++) a[i][nh] = a[nh][i] = true;
    nh = nr;
    nr += 2;
    n /= 2;
  } else {
    a[nr][nh] = a[nh][nr] = true;
    for (i = 0, j = nr, nr++; i < dpth - k; j = nr, nr++, i++)
      a[j][nr] = a[nr][j] = true;
    a[j][2] = a[2][j] = true;
    for (i = nr; i < nr + 2; i++) a[i][nh] = a[nh][i] = true;
    nh = nr;
    nr += 2;
    n = (n - 1) / 2;
  }
  for (k = 2; k <= dpth; k++) {
    if ((n & 1) == 0) {
      for (i = nh; i < nh + 2; i++)
        for (j = nr; j < nr + 2; j++) {
          a[i][j] = a[j][i] = true;
        }
      nh = nr;
      nr += 2;
      n /= 2;
    } else {
      for (i = nh; i < nh + 2; i++) a[i][nr] = a[nr][i] = true;
      for (i = 0, j = nr, nr++; i < dpth - k; j = nr, nr++, i++)
        a[j][nr] = a[nr][j] = true;
      a[j][2] = a[2][j] = true;
      for (i = nh; i < nh + 2; i++)
        for (j = nr; j < nr + 2; j++) {
          a[i][j] = a[j][i] = true;
        }
      nh = nr;
      nr += 2;
      n = (n - 1) / 2;
    }
  }
  for (i = nh; i <= nh + 2; i++) a[i][2] = a[2][i] = true;
  printf("%d\n", nr - 1);
  for (i = 1; i <= nr - 1; i++) {
    for (j = 1; j <= nr - 1; j++)
      if (a[i][j])
        printf("Y");
      else
        printf("N");
    printf("\n");
  }
  return 0;
}