A. Candy Bags
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald has n younger brothers and their number happens to be even. One day he bought n2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer k from 1 to n2 he has exactly one bag with k candies.

Help him give n bags of candies to each brother so that all brothers got the same number of candies.

Input

The single line contains a single integer n (n is even, 2 ≤ n ≤ 100) — the number of Gerald's brothers.

Output

Let's assume that Gerald indexes his brothers with numbers from 1 to n. You need to print n lines, on the i-th line print n integers — the numbers of candies in the bags for the i-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to n2. You can print the numbers in the lines in any order.

It is guaranteed that the solution exists at the given limits.

Examples
Input
2
Output
1 4
2 3
Note

The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

这题很惭愧,看了一个小时没有看出规律T T。

问了朋友,恍然大悟。

这种找规律的题我真的真的找不出来啊太笨了。

本题的规律就是:对角线上的和。

如下图:

n=3;

n=4;

图片来源:http://liyishuai.blog.ustc.edu.cn/candy-bags/

代码如下:

 1 #include<cstdio>
2 int main()
3 {
4 int n,i=1;
5 scanf("%d",&n);
6 for(int j=1;j<=n;j++)
7 {
8 i=j;
9 while(i<=n*n)
10 {
11 printf("%d ",i);
12 if(i%n==0) //在边界上 那下一步就要拐到下一行第一个
13 i+=1;
14 else
15 i+=n+1; //不在边界就斜着走到下一行
16 }
17 printf("\n");
18 }
19 return 0;
20 }

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