# 【数学规律】Sleepy Kaguya

Houraisan☆Kaguya is the princess who lives in Literally House of Eternity. However, she is very playful and often stays up late. This morning, her tutor, Eirin Yagokoro was going to teach her some knowledge about the Fibonacci sequence. Unfortunately, the poor princess was so sleepy in class that she fell asleep. Angry Eirin asked her to complete the following task:

This sequence can be described by following equations:
1.F=F=1

2.F[n]=F[n-1]+F[n-2]  (n>2)

Now, Kaguya is required to calculate F[k+1]*F[k+1]-F[k]*F[k+2] for each integer k that does not exceed 10^18.

Kaguya is so pathetic. You have an obligation to help her.
(I love Houraisan Kaguya forever!!!) image from pixiv,id=51208622

## 输入描述:

```Input
Only one integer k.```

## 输出描述:

```Output
Only one integer as the result which is equal to F[k+1]*F[k+1]-F[k]*F[k+2].```

`2`

`1`

## 说明

```F=1,F=2,F=3

2*2-1*3=1```

## 备注:

```0 < k ≤ 1^18

If necessary, please use %I64d instead of %lld when you use "scanf", or just use "cin" to get the cases.
```

1e18的复杂度肯定暴力常规做法不行。

G(k)=F[k+1]*F[k+1]-F[k]*F[k+2]
=F[k+1]*F[k+1]-F[k]*(F[k]+F[k+1])
=(F[k+1]-F[k])*F[k+1]-F[k]*F[k]
=F[k-1]*F[k+1]-F[k]*F[k]

```#include <iostream>
using namespace std;
int main()
{
long long k=0;
cin >> k;
if (k & 1)
cout << "-1" << endl;
else
cout << "1" << endl;
return 0;
}
```