CUIT ACM Personal Training 11.27(FM）A - Buggy Sor

A - Buggy Sorting

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of n integers a₁, a₂, ..., a_n in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number n and array a.

loop integer variable i from 1 to n - 1
    loop integer variable j from i to n - 1
        if (aj > aj + 1), then swap the values of elements aj and aj + 1

But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of n doesn't exist, print -1.

Input

You've got a single integer n (1 ≤ n ≤ 50) — the size of the sorted array.

Output

Print n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.

If there are several counter-examples, consisting of n numbers, you are allowed to print any of them.

Sample Input

Input

1

Output

-1

---

这个题的题意大概就说有人写了一个错误的冒泡排序，问你能否找出一个反例的输入，使得他的结果错误。
我们可以先来看看这个错误的冒泡排序，很容易发现，它每次的起始位置都在前移，也就是说比如一开始一串数字为4 3 2 1，第一次循环的时候起始位置是4，然后3与4进行交换，这串数字变成了3 4 2 1，然后依次交换之后变成了3 2 1 4。然而第二次循环的时候，他的起始位置数字不是3，而是2，那么3就永远不会被替换到了，所以最后结果肯定是错误的，所以可以看出，当我们的数字数量超过3个的时候，我们逆序排列输入，他的程序就会出错。而当只有一个数字和两个数字的时候，他的程序不进行交换，或者仅仅交换一次，那么我们就找不到反例，使得他的程序答案错误了。
所以AC代码如下；

#include<iostream>
using namespace std;

int main()
{
    int n;cin>>n;
    if(n==1||n==2) cout<<"-1"<<endl;
    else{for(int i=n;i>=1;i--) cout<<i<<" ";}
    cout<<endl;
    return 0;
}