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.*;
import java.io.*;

public class TaskB {
    private FastScanner in;

    public void solve() throws IOException {
        int k = in.nextInt();
        if (k == 1) {
            System.out.println("2");
            System.out.println("NY");
            System.out.println("YN");
            return;
        }
        int total = 500;
        final char[][] have = new char[total][total];
        for (int i = 0; i < total; i++) {
            for (int j = 0; j < total; j++) {
                have[i][j] = 'N';
            }
        }
        int left = 0, right = 1;
        int next = 2;
        for (int i = 1; i <= 30; i++) {
            have[left][next] = have[next][left] = 'Y';
            for (int j = 1; j <= i; j++) {
                for (int kk = left; kk < left + j; kk++) {
                    have[kk][next + j] = have[next + j][kk] = 'Y';
                }

            }
            left = next;
            right = next + (i + 1);
            next += (i + 1);
        }

        for (int i = 0; i <= 30; i++) {
            if (((1 << i) & k) != 0) {
                have[left + i + 1][1] = have[1][left + i + 1] = 'Y';
            }
        }
        System.out.println(total);
        for (int i = 0; i < total; i++) {
            for (int j = 0; j < total; j++) {
                System.out.print(have[i][j]);
            }
            System.out.println();
        }


    }

    public void run() {
        try {
            in = new FastScanner();
            solve();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }

    public static void main(String[] arg) {
        new TaskB().run();
    }
}