Codeforces Round #337 (Div. 2)C. Harmony Analysis

C. Harmony Analysis

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or  - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:

.

Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?

Input

The only line of the input contains a single integer k (0 ≤ k ≤ 9).

Output

Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to  - 1, and must be equal to ' + ' if it's equal to  + 1. It's guaranteed that the answer always exists.

If there are many correct answers, print any.

Examples

input

2

output

++**+*+*+++++**+

Note

Consider all scalar products in example:

• Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
• Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
• Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
• Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
• Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
• Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0

题意：求出2^k维的2^k个两两正交向量

/* ***********************************************Author       : rycCreated Time : 2016-08-20 SaturdayFile Name    : E:\acm\codeforces\337C.cppLanguage     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<queue>#include<vector>#include<stack>#include<map>using namespace std;typedef long long LL;typedef pair<int,int>pii;const int maxn=513;int num[maxn][maxn];void dabiao(){    num[1][1]=1;    for(int k=1;k<=256;k<<=1){        for(int i=1;i<=k;++i){            for(int j=1;j<=k;++j){                num[i+k][j]=num[i][j+k]=num[i][j];                num[i+k][j+k]=!num[i][j];            }        }    }}int main(){    dabiao();    int n;cin>>n;    for(int i=1;i<=1<<n;++i){        for(int j=1;j<=1<<n;++j){            if(num[i][j]){                printf("+");            }            else {                printf("*");            }        }        printf("\n");    }    return 0;}