Problem:
Score: 400 points

Problem Statement

Print a sequence a_1, a_2, ..., a_N whose length is N that satisfies the following conditions:

a_i (1 \leq i \leq N) is a prime number at most 55 555.

The values of a_1, a_2, ..., a_N are all different.

In every choice of five different integers from a_1, a_2, ..., a_N, the sum of those integers is a composite number.

If there are multiple such sequences, printing any of them is accepted.

Notes

An integer N not less than 2 is called a prime number if it cannot be divided evenly by any integers except 1 and N, and called a composite number otherwise.

Constraints

N is an integer between 5 and 55 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print N numbers a_1, a_2, a_3, ..., a_N in a line, with spaces in between.

Sample Input 1

5

Sample Output 1

3 5 7 11 31

Let us see if this output actually satisfies the conditions.

First, 3, 5, 7, 11 and 31 are all different, and all of them are prime numbers.

The only way to choose five among them is to choose all of them, whose sum is a_1+a_2+a_3+a_4+a_5=57, which is a composite number.

There are also other possible outputs, such as 2 3 5 7 13, 11 13 17 19 31 and 7 11 5 31 3.

Sample Input 2

6

Sample Output 2

2 3 5 7 11 13

2, 3, 5, 7, 11, 13 are all different prime numbers.

2+3+5+7+11=28 is a composite number.

2+3+5+7+13=30 is a composite number.

2+3+5+11+13=34 is a composite number.

2+3+7+11+13=36 is a composite number.

2+5+7+11+13=38 is a composite number.

3+5+7+11+13=39 is a composite number.

Thus, the sequence 2 3 5 7 11 13 satisfies the conditions.

Sample Input 3

8

Sample Output 3

2 5 7 13 19 37 67 79

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <iostream>
#include <algorithm>

typedef char                SINT8;
typedef unsigned char       UINT8;
typedef short               SINT16;
typedef unsigned short      UINT16;
typedef int                 SINT32;
typedef unsigned int        UINT32;
typedef long long           SINT64;
typedef unsigned long long  UINT64;
typedef double              DOUBLE;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))
#define rep(i,a,b) for(int (i)=int(a);(i)<int(b);(i)++)

#define INF = 1e9;
#define MOD = 1e9+7;
#define INF64 = 1e18;


using namespace std;

int main() {

	SINT32 N = 0;
	cin >> N;
	vector<SINT32> AR(55556);
	vector<SINT32> AR2(55556);
	vector<SINT32> AR3(55556);

	rep(i,2,55556) {
		AR[i] = i;
	}


	for(int i = 2; i < sqrt(55556); i++){
		if(AR[i] != 0) {
			for(int j = 2; i * j < 55556; j++) {
				AR[i * j] = 0;
			}
		}
	}
	AR2 = AR;

	sort(AR2.begin(),AR2.end());
	AR2.erase(std::unique(AR2.begin(), AR2.end()), AR2.end());




	AR3 = AR2;
	rep(i,0,AR3.size()) {
		AR3[i] %= 5;
	}

	
	SINT32 cnt = 0;
	while (N != 0) {
		if (AR3[cnt] == 1) {
			N--;
			cout << AR2[cnt] << " ";
		}
		cnt++;
	}

	return 0;
}

//	sort(DATA.begin(),DATA.end());
//	sort(DATA.begin(),DATA.end(),std::greater<SINT32>());
//	__gcd(ANS,DATA[i]);

//	// DATA.emplace_back(BUF);	//後ろに追加

//  DATA.erase(std::unique(DATA.begin(), DATA.end()), DATA.end());	//ソートしてから

//	DATA.insert(DATA.begin() + X, 0);	//X番目の要素に0を挿入


//	vector<vector<SINT32>> DP(N,vector<SINT32>(3));　//2次元配列

/*
	SINT32 M = 3;
	vector<pair<SINT32,SINT32>> DATA2(M);
	for (SINT32 i = 0; i < M; i++) {
		cin >> DATA2[i].first;
		cin >> DATA2[i].second;
	}
	sort(DATA2.begin(),DATA2.end());
	cout << DATA2[0].first << endl;
	cout << DATA2[0].second << endl;
*/

// 	vector<pair<int,pair<int,int>>> DATA(M);
//	cin >> DATA[i].first;
//	cin >> DATA[i].second.first;
//	cin >> DATA[i].second.second;


//		APOSI = lower_bound(ADATA.begin(),ADATA.end(),QDATA[i]) - ADATA.begin();
//		BPOSI = lower_bound(BDATA.begin(),BDATA.end(),QDATA[i]) - BDATA.begin();


/* 文字列回転
	string N;
	cin >> N;
	N = N[N.length()-1] + N.substr(0,N.length()-1);
*/

/* SET コンテナ
	set<SINT32> DATA;
	for (SINT32 i = 0; i < N; i++) {
		SINT32 buf;
		cin >> buf;
		DATA.insert(buf);
	}

	//全表示
    set<string>::iterator it; //  イテレータを用意
    for(it = DATA.begin(); it != DATA.end(); it++) {
        cout << *it << endl;
    }

	//N番目を一部表示
	set<string>::iterator it; //  イテレータを用意
	it = DATA.begin();
	rep (i,0,N) {
		it++;
	}
	cout << *it << endl;
*/

/* map
	map<string,SINT32> mp;
	SINT32 N = 0;
	SINT32 mx = 0;

	cin >> N;
	for (SINT32 i = 0; i < N; i++) {
		string s;
		cin >> s;
		mp[s]++;
	}
	for(auto it=mp.begin();it!=mp.end();it++) {
		mx=max(mx,it->second);
	}

*/