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 const nmax = 92;
bool v[5 + nmax][5 + nmax];
int main() {
  int k;
  cin >> k;
  v[1][3] = v[3][1] = 1;
  v[1][4] = v[4][1] = 1;
  int p = 63;
  for (int i = 2; i <= 30; i++) {
    v[i * 2 - 1][i * 2 + 1] = 1;
    v[i * 2 - 1][i * 2 + 2] = 1;
    v[i * 2][i * 2 + 1] = 1;
    v[i * 2][i * 2 + 2] = 1;
  }
  for (int i = 0; i < 29; i++) v[p + i][p + i + 1] = 1;
  v[p + 29][2] = 1;
  for (int i = 0; i < 30; i++) {
    if (0 < (k & (1 << i))) v[i * 2 + 4][p + i] = 1;
  }
  cout << nmax << '\n';
  for (int i = 1; i <= nmax; i++) {
    for (int j = 1; j <= nmax; j++)
      if (v[i][j] == 1 || v[j][i] == 1)
        cout << "Y";
      else
        cout << "N";
    cout << '\n';
  }
  return 0;
}