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.math.*;
import java.io.*;

public class B {


    char[][] fin = new char[320][320];
    void solve() throws Exception {
        for (char[] e: fin){
            Arrays.fill(e,  'N');
        }
        int n1 = ni();
        int len = Integer.toBinaryString(n1).length();
        addEdge(2, 103 + len - 1);
        addEdge(2, 203 + len - 1);

        addEdge(1, 3);
        for (int i = 0 ; i < len; i++){

            addEdge(3 + i, 3 + i + 1);

            addEdge(103 + i, 103 + i + 1);
            addEdge(103 + i, 203 + i + 1);

            addEdge(203 + i, 103 + i + 1);
            addEdge(203 + i, 203 + i + 1);

        }


        int ptr = 0;
//      px(Integer.toBinaryString(n1));
        for (char e: Integer.toBinaryString(n1).toCharArray()){
            if (e == '1'){
//              px(ptr + 3, ptr + 103);
                addEdge(ptr + 3, ptr + 103);
//              px(ptr + 102);
            }
            ptr++;
        }
        out.println(320);
        for (char[] e: fin){
            out.println(String.valueOf(e));
        }



    }

    void addEdge(int n1, int n2){
//      px(n1,n2);
        fin[n1 - 1][n2 - 1] = 'Y';
        fin[n2 - 1][n1 - 1] = 'Y';
    }

    public static void main(String[] args) throws Exception {new B();}

    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;

    String nx() throws Exception {
        for (; st == null || !st.hasMoreTokens(); ) {
            String k1 = in.readLine();
            if (k1 == null) { return k1; }
            st = new StringTokenizer(k1);
        }
        return st.nextToken();
    }

    int ni() throws Exception {
        return Integer.parseInt(nx());
    }

    long nl() throws Exception {
        return Long.parseLong(nx());
    }

    double nd() throws Exception {
        return Double.parseDouble(nx());
    }

    PrintWriter out = new PrintWriter(System.out);

    B() throws Exception {
        solve();
        out.flush();
    }

    static void px(Object... l1) {
        System.out.println(Arrays.deepToString(l1));
    }
}
