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: Wrong Answer)

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/
#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;

inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

#define si(x) scanf("%d",&x)
#define sii(x,y) scanf("%d %d",&x,&y)
#define siii(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define sl(x) scanf("%lld",&x)
#define sll(x,y) scanf("%lld %lld",&x,&y)
#define slll(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)
#define ss(ch) scanf("%s",ch)
#define pi(x) printf("%d",x)
#define pii(x,y) printf("%d %d",x,y)
#define piii(x,y,z) printf("%d %d %d",x,y,z)
#define pl(x) printf("%lld",x)
#define pll(x,y) printf("%lld %lld",x,y)
#define plll(x,y,z) printf("%lld %lld %lld",x,y,z)
#define ps(ch) printf("%s",ch)

/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
*/
///**
template < typename F, typename S >ostream& operator << ( ostream& os, const pair< F, S > & p ) {return os << "(" << p.first << ", " << p.second << ")";}
template < typename T >ostream &operator << ( ostream & os, const vector< T > &v ) {os << "{"; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << *it;} return os << "}";}
template < typename T >ostream &operator << ( ostream & os, const set< T > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin()) os << ", "; os << *it;} return os << "]";}
template < typename F, typename S >ostream &operator << ( ostream & os, const map< F, S > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ;} return os << "]";}
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)
clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
void faltu () { cerr << endl; }
template <typename T>void faltu( T a[], int n ) {for (int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;}
template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }

// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।
//*//**___________________________________________________**/
const int N = 1000006;

bool prime[N];

int main()
{
  //FASTIO
  /*
#ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  freopen("error.txt", "w", stderr);
#endif
//*/
  ll n;
  cin >> n;
  n--;
  cout << 2 << " ";
  for (int i = 4; i <= 55555; i += 2) prime[i] = 1;
  for (int i = 3; i < 55556; i += 2) {
    if (n <= 0) break;
    if (!prime[i]) {
      cout << i << " ";
      n--;
      for (int j = 2 * i; j < 55556; j += i)
        prime[j] = 1;
    }
  }
  return 0;
}