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;
int z[102], k, a[102], q, w, t, i, j, tot;
int f[1010][1010];
int main() {
  scanf("%d", &k);
  q = k;
  while (k > 0) {
    a[++tot] = k % 2;
    k = k / 2;
  }
  t = 1;
  w = 2;
  z[0] = 1;
  for (i = 1; i < tot; i++) {
    w++;
    f[t][w] = f[w][t] = 1;
    w++;
    f[t][w] = f[w][t] = 1;
    w++;
    z[i] = w;
    if (i == tot - 1) {
      w--;
      z[i] = 2;
      f[w][2] = f[2][w] = 1;
      f[w - 1][2] = f[2][w - 1] = 1;
      continue;
    }
    f[w - 2][w] = f[w][w - 2] = 1;
    f[w - 1][w] = f[w][w - 1] = 1;
    t = w;
  }
  for (i = tot - 1; i >= 1; i--)
    if (a[i]) {
      t = z[i - 1];
      for (j = i; j < tot; j++) {
        {
          int W = w + 1;
          f[t][W] = f[W][t] = 1;
          w++;
        }
        if (j == tot - 1)
          f[w][2] = f[2][w] = 1;
        else {
          int W = w + 1;
          f[w][W] = 1;
          w++;
          f[w][w - 1] = 1;
        }
        t = w;
      }
    }
  if (q == 1) f[1][2] = f[2][1] = 1;
  printf("%d\n", w);
  for (i = 1; i <= w; i++) {
    for (j = 1; j <= w; j++)
      if (f[i][j])
        putchar('Y');
      else
        putchar('N');
    putchar('\n');
  }
  return 0;
}