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.io.*;
import java.util.*;

public class p388b
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        System.out.println();
        int k = sc.nextInt();
        int n = 3*30+1+2*30+1;
        boolean g[][] = new boolean[n][n];
        int cur = 0;
        for (int i = 1; i <= 30; ++i) {
            int a = 2 * i;
            int b = 2 * i + 1;
            g[cur][a] = g[a][cur] = true;
            g[a][b] = g[b][a] = true;
            cur = b;
        }
        cur = 1;
        for (int i = 1; i <= 30; ++i) {
            int a = 59 + 3 * i;
            int b = 60 + 3 * i;
            int c = 61 + 3 * i;
            g[cur][a] = g[a][cur] = true;
            g[cur][b] = g[b][cur] = true;
            g[c][a] = g[a][c] = true;
            g[c][b] = g[b][c] = true;
            cur = c;
        }
        for (int bit = 0; bit <= 30; ++bit)
            if ((k & (1 << bit)) != 0) {
                int a;
                if (bit == 30) a = 0; else a = 61 - 2 * bit;
                int b;
                if (bit == 0) b = 1; else b = 61 + 3 * bit;
                g[a][b] = g[b][a] = true;
        }
        System.out.println(n);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) System.out.print(g[i][j] ? 'Y' : 'N');
            System.out.println();
        }
    }
}