codeforces-268B

B. Buttons

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Manao is trying to open a rather challenging lock. The lock has n buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.

Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.

Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.

Input

A single line contains integer n (1 ≤ n ≤ 2000) — the number of buttons the lock has.

Output

In a single line print the number of times Manao has to push a button in the worst-case scenario.

Examples

input

2

output

3

input

3

output

7

Note

Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.

这道题的意思大致为：输入数字n，表示n个按钮，需要按顺序全部按对才会成功，按错时会清空，输出最多按多少次按钮

5个按钮时，按钮次数参考下图

4　　3　　2　　1

3　　2　　1

2　　1

1

理解起来为第一个按钮不确定时需要按4次确定第一个按钮，然后按3次确定第二个按钮，期间第一个按钮按了3次，以此类推，最后争取的按钮按了n次。

#include<iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    int sum1 = 0, sum = 0;
    for (int i = 1;i < n;i++)
    {
        sum1 += i;
        sum += sum1;
    }
    sum += n;
    cout << sum << endl;
    return 0;
}