题目链接：http://codeforces.com/problemset/problem/715/A

ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) and '' (square root). Initially, the number 2 is displayed on the screen. There are n + 1 levels in the game and ZS the Coder start at the level 1.

When ZS the Coder is at level k, he can :

  1. Press the ' + ' button. This increases the number on the screen by exactly k. So, if the number on the screen was x, it becomes x + k.
  2. Press the '' button. Let the number on the screen be x. After pressing this button, the number becomes . After that, ZS the Coder levels up, so his current level becomes k + 1. This button can only be pressed when x is a perfect square, i.e. x = m2 for some positive integer m.

Additionally, after each move, if ZS the Coder is at level k, and the number on the screen is m, then m must be a multiple of k. Note that this condition is only checked after performing the press. For example, if ZS the Coder is at level 4 and current number is 100, he presses the '' button and the number turns into 10. Note that at this moment, 10 is not divisible by 4, but this press is still valid, because after it, ZS the Coder is at level 5, and 10 is divisible by 5.

ZS the Coder needs your help in beating the game — he wants to reach level n + 1. In other words, he needs to press the '' button n times. Help him determine the number of times he should press the ' + ' button before pressing the '' button at each level.

Please note that ZS the Coder wants to find just any sequence of presses allowing him to reach level n + 1, but not necessarily a sequence minimizing the number of presses.

Input

The first and only line of the input contains a single integer n (1 ≤ n ≤ 100 000), denoting that ZS the Coder wants to reach level n + 1.

Output

Print n non-negative integers, one per line. i-th of them should be equal to the number of times that ZS the Coder needs to press the ' + ' button before pressing the '' button at level i.

Each number in the output should not exceed 1018. However, the number on the screen can be greater than 1018.

It is guaranteed that at least one solution exists. If there are multiple solutions, print any of them.

Examples

Input

3

Output

14
16
46

Input

2

Output

999999999999999998
44500000000

Input

4

Output

2
17
46
97

Note

In the first sample case:

On the first level, ZS the Coder pressed the ' + ' button 14 times (and the number on screen is initially 2), so the number became 2 + 14·1 = 16. Then, ZS the Coder pressed the '' button, and the number became .

After that, on the second level, ZS pressed the ' + ' button 16 times, so the number becomes 4 + 16·2 = 36. Then, ZS pressed the '' button, levelling up and changing the number into .

After that, on the third level, ZS pressed the ' + ' button 46 times, so the number becomes 6 + 46·3 = 144. Then, ZS pressed the '' button, levelling up and changing the number into .

Note that 12 is indeed divisible by 4, so ZS the Coder can reach level 4.

Also, note that pressing the ' + ' button 10 times on the third level before levelling up does not work, because the number becomes 6 + 10·3 = 36, and when the '' button is pressed, the number becomes  and ZS the Coder is at Level 4. However, 6 is not divisible by 4 now, so this is not a valid solution.

In the second sample case:

On the first level, ZS the Coder pressed the ' + ' button 999999999999999998 times (and the number on screen is initially 2), so the number became 2 + 999999999999999998·1 = 1018. Then, ZS the Coder pressed the '' button, and the number became .

After that, on the second level, ZS pressed the ' + ' button 44500000000 times, so the number becomes 109 + 44500000000·2 = 9·1010. Then, ZS pressed the '' button, levelling up and changing the number into .

Note that 300000 is a multiple of 3, so ZS the Coder can reach level 3.

题目大意： 
你在玩一个游戏，游戏机上有一个屏幕，屏幕上有一个数字。还有两个按钮：加号和根号。 
初始条件下，屏幕数字为2，你在第一关。 
两个按钮的作用如下： 
加号：若屏幕上的数字为x，你在第k关，则变成x+k。 
根号：若屏幕上的数字为x，你在第k关，则必须在x是完全平方数，且开完平方后的数字是x+1的倍数时才可以按下，变成x的平方根，且进入第k+1关。

输入n，求要想过第n关，每一关需按多少+才按根号？

输出如果有多种情况，输出任意一种就可以。
 

首先暴力模拟肯定过不了,那么我们考虑找一下规律。

1.在第k关，每一次都加k,而初始值也是k的倍数，也就是说第k关的所有数都是k的倍数

2.若在第k关想开根号，则值必须是(k+1)*(k+1)的倍数

综合1,2，我们可以构造第k关最后一个数为k*k*(k+1)*(k+1),所以对应的第k关开始的那个数为(k-1)*k   (第一关除外）

所以第k关按加号的个数为[k*k*(k+1)*(k+1)-(k-1)*k]/k   (因为每次加k所以要除以k），

即k*(k+1)*(k+1)-(k-1).

特例：

第一关的初始值是2，并不是（1-1）*1=0，而第二关的初始值是2，所以第一关的最后一个值应该为4，需要加2次

 

#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
int main() {
	int n;
	scanf("%I64d",&n);
    for(ll i=1;i<=n;i++){
    	if(i==1) 
		printf("2\n");
    	else
    	printf("%I64d\n",i*(i+1)*(i+1)-(i-1);
	}
	return 0;
}