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 Bdiv1 {

	public static void main(String[] args) {
		int m = 1000; 
		char[][] res = new char[m][m];
		for(int i = 0; i < m; i++) {
			Arrays.fill(res[i], 'N');
		}
		int n = 30;
		int[] starts = new int[n];
		for(int i = 1; i <= n; i++) {
			res[i][i+1] = res[i+1][i] = 'Y';
		}
		starts[0] = n+1;
		int a = n+2, pa = -1;
		for(int s = 1; s < n; s++) {
			if(s == 1) {
				res[a][1] = res[1][a] = 'Y';
				res[a+1][1] = res[1][a+1] = 'Y';
			}
			else {
				res[a][pa] = res[pa][a] = 'Y';
				res[a][pa+1] = res[pa+1][a] = 'Y';
				res[a+1][pa] = res[pa][a+1] = 'Y';
				res[a+1][pa+1] = res[pa+1][a+1] = 'Y';
			}
			res[a][a+2] = res[a+2][a] = 'Y';
			res[a+1][a+2] = res[a+2][a+1] = 'Y';
			int c = a+2;
			for(int i = 0; i < n-s-1; i++) {
				res[c][c+1] = res[c+1][c] = 'Y'; c++;
			}
			starts[s] = c; pa = a; a = c+1;
		}
//		ArrayList<Integer>[] g = new ArrayList[M];
//		for(int i = 0; i < M; i++) {
//			g[i] = new ArrayList<>();
//		}
//		for(int i = 0; i < M; i++) {
//			for(int j = 0; j < M; j++) {
//				if(res[i][j] == 'Y') {
//					g[i].add(j);
//				}
//			}
//		}
//		
		FastScanner sc = new FastScanner();
		int k = sc.nextInt();
		for(int s = 0; s < n; s++) {
			if((k & (1 << s)) > 0) {
				res[0][starts[s]] = res[starts[s]][0] = 'Y'; 
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.println("1000");
		for(int i = 0; i < m; i++) {
			pw.println(res[i]);
		}
		pw.flush();
	}
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }

}
