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 JAVA):

Solution:
import java.util.*;

public class B{
	static int ind = 2;
	static boolean grid[][];
	static int count(){
		int length[][] = new int[grid.length][grid.length];
		int ret[][] = new int[grid.length][grid.length];
		for(int i = 0; i < grid.length; i++)
			for(int j = 0; j < grid.length; j++)
			{
				if(grid[i][j])
				{
					length[i][j] = 1;
					ret[i][j] = 1;
				}
				else length[i][j] = grid.length*2;
			}

		for(int k = 0; k < grid.length; k++)
			for(int i = 0 ; i < grid.length; i++)
				for(int j = 0; j < grid.length; j++)
				{
					if(length[i][k] + length[k][j] < length[i][j]){
						length[i][j] = length[i][k] + length[k][j];
						ret[i][j] = ret[i][k]*ret[k][j];
					}
					else if (length[i][k] + length[k][j] == length[i][j]){
						ret[i][j] += ret[i][k]*ret[k][j];
					}
				}
		return ret[0][1];
	}
	static int[] next()
	{	
		int ret[] = new int[3];
		for(int i = 0; i < 3; i++)
			ret[i] = ind++;
		return ret;
	}
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String str = Integer.toBinaryString(n);
		// System.out.println(str);
		int temp = n;
		int count = 0;
		while(temp > 1){
			temp/=2;
			count++;
		}

		int size = 3*count + 2;
		System.out.println(size);
		grid = new boolean[size][size];

		if(size == 2){
			System.out.println("NY");
			System.out.println("YN");
		}
		else{
			int s = 0;
			int e = 1;
			int curr[] = next();

			for(int i = 0; i < 3; i++)
			{
				grid[s][curr[i]] = true;
				grid[curr[i]][s] = true;
			}
			for(int i = count-1; i > 0; i--){
				int next[] = next();
				grid[curr[0]][next[0]] = grid[next[0]][curr[0]] = true;
				grid[curr[1]][next[1]] = grid[next[1]][curr[1]] = true;
				grid[curr[1]][next[2]] = grid[next[2]][curr[1]] = true;
				grid[curr[2]][next[1]] = grid[next[1]][curr[2]] = true;
				grid[curr[2]][next[2]] = grid[next[2]][curr[2]] = true;
				
				if((n&(1<<i)) != 0){
					grid[curr[0]][next[1]] = grid[next[1]][curr[0]] = true;
					grid[curr[0]][next[2]] = grid[next[2]][curr[0]] = true;
				}
				curr = next;
			}
			if(n%2 != 0)
			{
				grid[curr[0]][e] = grid[e][curr[0]] = true;			
			}
			grid[curr[1]][e] = grid[e][curr[1]] = true;
			grid[curr[2]][e] = grid[e][curr[2]] = true;
			for(int i = 0 ;i < size; i++)
			{
				StringBuilder sb = new StringBuilder();
				for(int j = 0; j < size; j++)
					if(grid[i][j])
						sb.append("Y");
					else sb.append("N");
				System.out.println(sb.toString());
			}


			// System.out.println(count());
		}
	}
}