Codeforces 388B Fox and Minimal path（构造最短路条数为N的图）

B. 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 withn 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 tok?

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 exactlyk shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrixG with n rows andn columns must follow. Each element of the matrix must be 'N' or 'Y'. IfG_ij is 'Y', then graphG has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 ton.

The graph must be undirected and simple: G_ii = 'N' andG_ij = G_ji 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

4NNYYNNYYYYNNYYNN

Input

9

Output

8NNYYYNNNNNNNNYYYYNNNNYYYYNNNNYYYYNNNNYYYNYYYYNNNNYYYYNNNNYYYYNNN

Input

1

Output

2NYYN

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.

题目大意：

    输入一个数字N，构造一个最短路径数为N的图，以邻接矩阵的形式输出。

解题思路：

    开始想的是因数分解，不过这样对于非常大的质数会超过1000个点的限制。后来在吃饭的时候灵光一闪想到的正确的方法(￣▽￣)

    任何一个数都可以用2进制表示，而且1e9也才二十几位而已。所以我们就可以把N拆成2^a+2^b+2^c…。于是我们就可以构造一条主链，表示主要路径，一条辅链，表示2^x。如图

    因为主链已经表示了1，所以我们要把N-=1，然后对N进行转化为二进制数，如果第i位为1就把主链第i+1层和辅链第i层的一个节点连接，相当于增加了2^I条路径。这样就可以得到满足题目要求的图了。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1000+3;int N,size;//需要路径数、图的层数bool G[maxn][maxn];//邻接矩阵void add_edge(int a,int b)//加边{    G[a][b]=G[b][a]=true;}int main(){    scanf("%d",&N);    --N;    int tmp=N;    while(tmp>0)    {        ++size;        tmp/=2;    }    if(size==0)//N==1时特判    {        printf("2\nNY\nYN\n");        return 0;    }    //构造主链和辅链    add_edge(1,3);    add_edge(1,4);    add_edge(1,5);    add_edge(size*3,2);    for(int i=1;i<size;++i)    {        add_edge(i*3,(i+1)*3);        add_edge(i*3+1,(i+1)*3+1);        add_edge(i*3+2,(i+1)*3+2);        add_edge(i*3+1,(i+1)*3+2);        add_edge(i*3+2,(i+1)*3+1);    }    int now=0;    while(N>0)    {        ++now;        if(N&1)//增加2^now条路径        {            if(now==size)                add_edge(2,3*now+1);            else add_edge(3*(now+1),3*now+1);        }        N/=2;    }    size=size*3+2;    printf("%d\n",size);    for(int i=1;i<=size;++i)//输出邻接矩阵    {        for(int j=1;j<=size;++j)            putchar(G[i][j]?'Y':'N');        putchar('\n');    }        return 0;}