Fox and Minimal path CodeForces

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.

本地的标准写法构造三进制

由于的是用二进制会有极限的数据卡你图的大小

暂时先把re代码发上来吧

留坑

#include<cstdio>
#include<map>
#include<cstring>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
bool mp[1005][1005];
int highbits(int x)
{
    int cnt=0;
    while(x)
    {
        cnt++;
        x=x>>1;
    }
    return cnt-1;
}
int main()
{
    memset(mp,0,sizeof(mp));
    int n;
    scanf("%d",&n);
    int tmp;//取临时位
    int pos=3;//记录矩阵目前状态
    int bits=highbits(n);
    tmp=n;
    if(bits==0)
    {
        mp[1][2]=1;
    }
    for(int i=0; i<=bits; i++)
    {
        int temp=(tmp&1);
        //cout<<"tmp"<<tmp<<" "<< temp <<endl;
        //cout<<i<<endl;
        if(temp==1)
        {
            //cout<<i<<endl;
            int flag=0;
            for(int j=1; j<=bits; j++)
            {

                if(j<=i)//构成二进制环形
                {
                    if(j==1)
                    {
                        mp[1][pos]=1;
                        pos++;
                        mp[1][pos]=1;
                    }
                    else
                    {
                        mp[pos-2][pos]=1;
                        mp[pos-1][pos]=1;
                        pos++;
                        mp[pos-3][pos]=1;
                        mp[pos-2][pos]=1;
                    }
                    flag=1;
                }
                else
                {
                    if(j==1)
                    {
                        mp[1][pos]=1;
                    }
                    else
                    {
                        if(flag==1)
                        {
                            mp[pos-1][pos]=1;
                            mp[pos-2][pos]=1;
                        }
                        else
                        mp[pos-1][pos]=1;
                    }
                    flag=0;
                }
                pos++;
            }
            if(i==bits)
            {
                mp[2][pos-2]=1;
                mp[2][pos-1]=1;
            }
            else
            {
                mp[2][pos-1]=1;
            }
        }
        tmp=tmp>>1;
    }
    for(int i=1; i<=1000; i++)
    {
        for(int j=1; j<=1000; j++)
        {
            if(i==j) mp[i][j]=0;
            if(i>j) mp[i][j]=mp[j][i];
        }
    }
//    for(int i=0;i<=9;i++)
//        cout<<i;
//    cout<<endl;
    printf("1000
");
    for(int i=1; i<=1000; i++)
    {
        //cout<<i%10;
        for(int j=1; j<=1000; j++)
        {
            if(mp[i][j]==1)
            {
                printf("Y");
            }
            else
            {
                printf("N");
            }
        }
        printf("
");
    }
}
//caowenbo