Codeforces389D(SummerTrainingDay01-J)

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

output

4
NNYY
NNYY
YYNN
YYNN

input

output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

input

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。

 1 //2017-10-15
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 1100;
10 bool a[N][N];
11 
12 int main()
13 {
14     long long k;
15     while(cin>>k){
16         int n = 0;
17         while((1<<n) <= k)
18               n++;
19         if(n)n--;
20         int step = n*2+1;
21         int num = (1<<n);
22         memset(a, 0, sizeof(a));
23         n = 3*n+3;
24         a[1][3] = 1;
25         for(int i = 4; i <= n; i++){
26             if(i%3 == 0)a[i][i-1] = a[i][i-2] = 1;
27             else if((i-1)%3 == 0)a[i][i-1] = 1;
28             else if((i-2)%3 == 0)a[i][i-2] = 1;
29         }
30         a[n][2] = 1;
31         for(int i = n+1; i < n+step; i++)
32               a[i][i+1] = 1;
33         a[n+step-1][2] = 1;
34         int tmp = k - num, ptr = 1;
35         while(tmp){
36             int fg = (1&tmp);
37             tmp>>=1;
38             if(fg)a[(ptr+1)/2*3][ptr+n] = 1;
39             ptr+=2;
40         }
41         if(k != num)n+=(step-1);
42         cout<<n<<endl;
43         for(int i = 1; i <= n; i++){
44               for(int j = 1; j <= n; j++)
45                 if(a[i][j] || a[j][i])
46                   cout<<'Y';
47                 else cout<<'N';
48             cout<<endl;
49         }
50     }
51 
52     return 0;
53 }