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.io.PrintWriter;
import java.util.Scanner;

public class D {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int k = in.nextInt();
        if (k == 1) {
            System.out.println(2);
            System.out.println("NY\nYN");
            System.exit(0);
        }
        int l = 1;
        for (int i = 1; i < 31; i++)
            if ((k & (1 << i)) != 0)
                l = i;
        char[][] res = new char[1000][1000];
        int st = 2;
        int a = st++;
        int b = st++;
        res[0][a] = res[a][0] = 'Y';
        res[0][b] = res[b][0] = 'Y';
        for (int j = 1; j < l; j++) {
            int newA = st++;
            int newB = st++;
            res[a][newA] = res[a][newB] = res[b][newA] = res[b][newB] = 'Y';
            res[newA][a] = res[newB][a] = res[newA][b] = res[newB][b] = 'Y';
            a = newA;
            b = newB;
        }
        res[a][1] = res[b][1] = 'Y';
        res[1][a] = res[1][b] = 'Y';

        for (int i = 0; i < 31; i++)
            if ((k & (1 << i)) != 0) {
                int left = l - i;
                int temp = 0;
                for (int j = 0; j < i; j++)
                    temp += 2;
                if (left > 0) {
                    if (temp == 0)
                        res[temp][st] = res[st][temp] = 'Y';
                    else
                        res[temp][st] = res[st][temp] = res[st][temp + 1] = res[temp + 1][st] = 'Y';
                    a = st++;
                    for (int j = 1; j < left; j++) {
                        res[a][st] = res[st][a] = 'Y';
                        a = st++;
                    }
                    res[a][1] = res[1][a] = 'Y';
                }
            }
        out.println(st);
        for (int i = 0; i < st; i++) {
            for (int j = 0; j < st; j++)
                out.print(res[i][j] == 'Y' ? res[i][j] : 'N');
            out.println();
        }
        out.flush();
        out.close();
    }
}