Codeforces Round #259 (Div. 2) A. Little Pony and Crystal Mine

A. Little Pony and Crystal Mine

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n × n matrix with a diamond inscribed into it.

You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.

Input

The only line contains an integer n (3 ≤ n ≤ 101; n is odd).

Output

Output a crystal of size n.

Examples

Input

3

Output

*D*
DDD
*D*

Input

5

Output

**D**
*DDD*
DDDDD
*DDD*
**D**

Input

7

Output

***D***
**DDD**
*DDDDD*
DDDDDDD
*DDDDD*
**DDD**
***D***







ac代码：
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int l=n/2+1,r=n/2+1;
		for(int i=1;i<=n;i++)
		{
			int j=1;
			while(j<l)
			printf("*"),j++;
			while(j<=r)
			printf("D"),j++;
			while(j<=n)
			printf("*"),j++;
			printf("\n");
			if(i>=n/2+1)
			l++,r--;
			else
			l--,r++;
		}
	}
	return 0;
}