Codeforces389D(SummerTrainingDay01-J)

D. Fox and Minimal path

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with nvertexes. 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.

题意：求正好有k条最短路的图的邻接矩阵。

思路：二进制思想构图。

 //2017-10-15
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;
 bool a[N][N];

 int main()
 {
     long long k;
     while(cin>>k){
         int n = ;
         while((<<n) <= k)
               n++;
         if(n)n--;
         int step = n*+;
         int num = (<<n);
         memset(a, , sizeof(a));
         n = *n+;
         a[][] = ;
         for(int i = ; i <= n; i++){
             if(i% == )a[i][i-] = a[i][i-] = ;
             else if((i-)% == )a[i][i-] = ;
             else if((i-)% == )a[i][i-] = ;
         }
         a[n][] = ;
         for(int i = n+; i < n+step; i++)
               a[i][i+] = ;
         a[n+step-][] = ;
         int tmp = k - num, ptr = ;
         while(tmp){
             int fg = (&tmp);
             tmp>>=;
             if(fg)a[(ptr+)/*][ptr+n] = ;
             ptr+=;
         }
         if(k != num)n+=(step-);
         cout<<n<<endl;
         for(int i = ; i <= n; i++){
               for(int j = ; j <= n; j++)
                 if(a[i][j] || a[j][i])
                   cout<<'Y';
                 else cout<<'N';
             cout<<endl;
         }
     }

     return ;
 }