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 = 1100;
bool a[N][N];

int main()
{
    long long k;
    while(cin>>k){
        int n = 0;
        while((1<<n) <= k)
              n++;
        if(n)n--;
        int step = n*2+1;
        int num = (1<<n);
        memset(a, 0, sizeof(a));
        n = 3*n+3;
        a[1][3] = 1;
        for(int i = 4; i <= n; i++){
            if(i%3 == 0)a[i][i-1] = a[i][i-2] = 1;
            else if((i-1)%3 == 0)a[i][i-1] = 1;
            else if((i-2)%3 == 0)a[i][i-2] = 1;
        }
        a[n][2] = 1;
        for(int i = n+1; i < n+step; i++)
              a[i][i+1] = 1;
        a[n+step-1][2] = 1;
        int tmp = k - num, ptr = 1;
        while(tmp){
            int fg = (1&tmp);
            tmp>>=1;
            if(fg)a[(ptr+1)/2*3][ptr+n] = 1;
            ptr+=2;
        }
        if(k != num)n+=(step-1);
        cout<<n<<endl;
        for(int i = 1; i <= n; i++){
              for(int j = 1; j <= n; j++)
                if(a[i][j] || a[j][i])
                  cout<<'Y';
                else cout<<'N';
            cout<<endl;
        }
    }

    return 0;
}