Codeforces Beta Round #61 (Div. 2) D. Petya and His Friends 想法

D. Petya and His Friends

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya has a birthday soon. Due this wonderful event, Petya's friends decided to give him sweets. The total number of Petya's friends equals to n.

Let us remind you the definition of the greatest common divisor: GCD(a₁, ..., a_k) = d, where d represents such a maximal positive number that each a_i (1 ≤ i ≤ k) is evenly divisible by d. At that, we assume that all a_i's are greater than zero.

Knowing that Petya is keen on programming, his friends has agreed beforehand that the 1-st friend gives a₁ sweets, the 2-nd one gives a₂ sweets, ..., the n-th one gives a_n sweets. At the same time, for any i and j (1 ≤ i, j ≤ n) they want the GCD(a_i, a_j) not to be equal to 1. However, they also want the following condition to be satisfied: GCD(a₁, a₂, ..., a_n) = 1. One more: all the a_i should be distinct.

Help the friends to choose the suitable numbers a₁, ..., a_n.

Input

The first line contains an integer n (2 ≤ n ≤ 50).

Output

If there is no answer, print "-1" without quotes. Otherwise print a set of n distinct positive numbers a₁, a₂, ..., a_n. Each line must contain one number. Each number must consist of not more than 100 digits, and must not contain any leading zeros. If there are several solutions to that problem, print any of them.

Do not forget, please, that all of the following conditions must be true:

• For every i and j (1 ≤ i, j ≤ n): GCD(a_i, a_j) ≠ 1
• GCD(a₁, a₂, ..., a_n) = 1
• For every i and j (1 ≤ i, j ≤ n, i ≠ j): a_i ≠ a_j

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3

Output

99
55
11115

Input

4

Output

385
360
792
8360
题意：满足这三个条件，否则输出-1；

• For every i and j (1 ≤ i, j ≤ n): GCD(a_i, a_j) ≠ 1
• GCD(a₁, a₂, ..., a_n) = 1
• For every i and j (1 ≤ i, j ≤ n, i ≠ j): a_i ≠ a_j

　思路：2显然-1；

　　　　先找到一个3满足的条件，然后输出一个数的倍数即可；注意不要重复；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 1e-13
const int N=1e4+10,M=1e6+50000,inf=1e9+10,mod=1000000007;
int main()
{
    int x,y,i,z,t;
    scanf("%d",&x);
    if(x==2)
    printf("-1");
    else
    {
        printf("6
10
15
");
        for(int i=0,t=2;i<x-3;i++,t++)
        printf("%d
",6*t);
    }
    return 0;
}