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 ntimes. 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.

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.

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.

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.

题意：当前数字a[k]（a[k]是k的倍数）要么+k,要么开根得到a[i+1]（必须完全平方根），问每次得到下一个数要几次加

想了个暴力，枚举c当前数是c*c*(k+1)*(k+1)，找满足a[k]+k*d的

然而正解是构造，好神奇

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官方题解：

Firstly, let ai(1 ≤ i ≤ n) be the number on the screen before we level up from level i to i + 1. Thus, we require all the ais to be perfect square and additionally to reach the next ai via pressing the plus button, we require  and  for all 1 ≤ i < n. Additionally, we also require ai to be a multiple of i. Thus, we just need to construct a sequence of such integers so that the output numbers does not exceed the limit 1018.

There are many ways to do this. The third sample actually gave a large hint on my approach. If you were to find the values of ai from the second sample, you'll realize that it is equal to 4, 36, 144, 400. You can try to find the pattern from here. My approach is to use ai = [i(i + 1)]2. Clearly, it is a perfect square for all 1 ≤ i ≤ n and when n = 100000, the output values can be checked to be less than 1018

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which is a multiple of i + 1, and  is also a multiple of i + 1.

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a[i]＝i*i*(i+1)*(i+1)

因为a[i]是i的倍数又是(i+1)平方的倍数并且a[i]<a[i+1]

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<ctime>

using namespace std;

typedef long long ll;

int n,k=;

ll x=;

int main(int argc, const char * argv[]) {

    scanf("%d",&n);

    printf("2\n");

    for(int k=;k<=n;k++){

        printf("%I64d\n",(ll)k*(k+)*(k+)-(k-));

    }

    return ;

}