time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

【题解】



规律题;

k和k-1的答案存在如下关系;



上图中的方框表示k-1时的答案;

把它按照上述方式复制3份，第4份则在原来的基础上取反;(用1表示+，0表示减);

比如样例输入

++**

+*+*

++++

+**+

//->

1100

1010

1111

1001

//->相同的3份取反的一份

1100 1100

1010 1010

1111 1111

1001 1001

1100 0011

1010 0101

1111 0000

1001 0110

/*而这正是k=3时的答案;右下角那个

    上面两个方框是肯定满足的;

    为了让下面两个方框在乘的时候也满足;

    相当于左边取A，右边取它的相反数-A;这样一减就是0;

    如果一个答案符合要求则全部取反还是能符合要求的;

    所以右下角取反不会影响下面两个的答案正确性;

    然后又能让上面两个方框的乘下面两个方框的向量的时候积为0；所以是符合要求的;

*/

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <set>

#include <map>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <queue>

#include <vector>

#include <stack>

#include <string>

#define lson L,m,rt<<1

#define rson m+1,R,rt<<1|1

#define LL long long

using namespace std;

const int MAXN = 1000;

const int dx[5] = {0,1,-1,0,0};

const int dy[5] = {0,0,0,-1,1};

const double pi = acos(-1.0);

struct abc

{

    int a[1000][1000];

};

int k;

abc ans[10];

void input_LL(LL &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t) && t!='-') t = getchar();

    LL sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

void input_int(int &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)&&t!='-') t = getchar();

    int sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

int main()

{

   // freopen("F:\\rush.txt","r",stdin);

    ans[0].a[1][1] = 0;

    for (k= 1;k <= 9;k++)

    {

        for (int i = 1;i <= 1<<(k-1);i++)

            for (int j = 1;j <= 1<<(k-1);j++)

            {

                ans[k].a[i][j] = ans[k-1].a[i][j];

                ans[k].a[i+(1<<(k-1))][j] = ans[k-1].a[i][j];

                ans[k].a[i][j+(1<<(k-1))] = ans[k].a[i][j];

                ans[k].a[i+(1<<(k-1))][j+(1<<(k-1))] = !ans[k].a[i][j];

            }

    }

    input_int(k);

    for (int i = 1;i <= 1<<k;i++)

    {

        for (int j = 1;j <= 1<<k;j++)

            if (ans[k].a[i][j])

                putchar('+');

            else

                putchar('*');

        puts("");

    }

    return 0;

}