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 f[1005][1005];
int main() {
  long long k, i, j, lj1, lj2;
  cin >> k;
  lj1 = 0;
  lj2 = 1;
  j = 3;
  i = 30;
  while (k > 0) {
    if (k % 2 == 1) {
      f[lj1][j] = true;
      f[lj2][j] = true;
      f[j][lj1] = true;
      f[j][lj2] = true;
      j++;
      for (int a = i; a > 0; --a) {
        f[j - 1][j] = true;
        f[j][j - 1] = true;
        j++;
      }
      f[j - 1][2] = true;
      f[2][j - 1] = true;
    }
    f[lj1][j] = true;
    f[lj1][j + 1] = true;
    f[lj2][j] = true;
    f[lj2][j + 1] = true;
    f[j][lj1] = true;
    f[j][lj2] = true;
    f[j + 1][lj1] = true;
    f[j + 1][lj2] = true;
    lj1 = j;
    lj2 = j + 1;
    j += 2;
    k = k / 2;
    i--;
  }
  cout << j - 1 << endl;
  for (int a = 1; a < j; ++a) {
    for (int b = 1; b < j; ++b)
      if (f[a][b])
        cout << "Y";
      else
        cout << "N";
    cout << endl;
  }
  return 0;
}