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.
import java.util.*;
import java.io.*;
public class FoxAndMinimalPath
{
	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int k = in.nextInt();
		
		int n = 154;
		boolean[][] adj = new boolean[n][n];
		for(int x = 0; x < 30; x++)
		{
			int cur = 5 * x + 2;
			add(cur, cur + 2, adj);
			add(cur + 2, cur + 5, adj);
			add(cur + 1, cur + 3, adj);
			add(cur + 1, cur + 4, adj);
			add(cur + 3, cur + 6, adj);
			add(cur + 4, cur + 6, adj);
		}
		
		int cur = 2;
		while(k > 0)
		{
			if(k % 2 == 1)
			{
				add(cur, cur + 1, adj);
			}
			
			cur += 5;
			k /= 2;
		}
		
		add(0, 3, adj);
		add(n - 2, 1, adj);
		
		System.out.println(n);
		for(int y = 0; y < adj.length; y++)
		{
			for(int z = 0; z < adj[0].length; z++)
			{
				if(adj[y][z])
				{
					System.out.print("Y");
				}
				else
				{
					System.out.print("N");
				}
			}
			System.out.println();
		}
	}
	
	public static void add(int u, int v, boolean[][] adj)
	{
		adj[u][v] = true;
		adj[v][u] = true;
	}
}
