Problem: Given n, find n consecutive positive integers. However, all numbers must have divisors other than 1 and the number itself.

Hint

In Sample Output 2, 8, 9 and 10 are selected as three consecutive integers.
The second and third lines output 3, as a divisor of 9, and the fourth line outputs 5, as a divisor of 10.



Input

The input is given in the following format.


n


The input meets the following constraints.
1 ≤ n ≤ 1,500

Output

Print the smallest of the n consecutive positive integers you choose on the first line.
Output the divisor for each value from the 2nd line to the n + 1 line.
Any value may be output unless the divisor is 1 or the number itself.
Let x be the number output on the first line, and output the divisor of x + i-2 on the i line.

The output value must not exceed 5,000 digits.

Examples

Input

2


Output

8
2
3


Input

3


Output

8
2
3
5

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <vector>
#include <list>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <iterator>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <stack>
#include <climits>
#include <deque>
#include <bitset>
#include <cassert>
#include <ctime>
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
const int dy[]={-1,0,1,0},dx[]={0,1,0,-1};
// adjust problem by problem
const double EPS=1e-8;
const double PI=acos(-1.0);
#ifdef __GNUC__
int popcount(int n){return __builtin_popcount(n);}
int popcount(ll n){return __builtin_popcountll(n);}
#endif
#ifndef __GNUC__
template<class T> int popcount(T n){int cnt=0;while(n){if(n%2)cnt++;n/=2;}return cnt;}
#endif
template<class T>int SIZE(T a){return a.size();}
template<class T>string IntToString(T num){string res;stringstream ss;ss<<num;return ss.str();}
template<class T>T StringToInt(string str){T res=0;for(int i=0;i<SIZE(str);i++)res=(res*10+str[i]-'0');return res;}
template<class T>T gcd(T a,T b){if(b==0)return a;return gcd(b,a%b);}
template<class T>T lcm(T a,T b){return a/gcd(a,b)*b;}
bool EQ(double a,double b){return abs(a-b)<EPS;}
void fastStream(){cin.tie(0);std::ios_base::sync_with_stdio(0);}
vector<string> split(string str,char del){
  vector<string> res;
  for(int i=0,s=0;i<SIZE(str);i++){
    if(str[i]==del){if(i-s!=0)res.push_back(str.substr(s,i-s));s=i+1;}
    else if(i==SIZE(str)-1){res.push_back(str.substr(s));}
  }
  return res;
}

const int MAX_PRIME = 10000000;
int prime[MAX_PRIME+1];
bool isPrime[MAX_PRIME+1];
int p;
void erats(){
  // primeリストを初期化
  fill(isPrime,isPrime+MAX_PRIME,true);
  p=0;
  isPrime[0]=isPrime[1]=false;
  for(int i=2;i<=MAX_PRIME;i++){
    if(isPrime[i]){
      prime[p++]=i;
      // ふるいにかける
      for(int j=2*i;j<=MAX_PRIME;j+=i)isPrime[j] = false;
    }
  }
}


string add(string str1,string str2){
  string str;
  int carry = 0;
  for(int i = 0; i < max(str1.size(),str2.size()); i++){
    if(i >= str1.size()){
      int ret = carry + str2[str2.size()-1-i] - '0';
      carry = ret / 10;
      str += (ret % 10) + '0';
    }
    else if(i >= str2.size()){
      int ret = carry + str1[str1.size()-1-i] - '0';
      carry = ret / 10;
      str += (ret % 10) + '0';
    }
    else{
      int ret = str1[str1.size()-1-i] + str2[str2.size()-1-i] - '0' - '0' + carry;
      carry = ret / 10;
      str += (ret % 10) + '0';
    }
  }
  str+=(carry+'0');
  // 破壊的変更
  reverse(str.begin(),str.end());
  // 先頭の0を消す
  int idx;
  for(int i = 0; i < str.size(); i++){
    idx=i;
    if(str[i]!='0'||str.size()-1==i)
      break;
  }
  return str.substr(idx);
}
string mul(string str1,string str2){
  int base=0;
  string ret="0";
  for(int j = str2.size()-1; j >= 0; j--){
    string str;
    int carry = 0;
    for(int i = str1.size()-1; i >=0; i--){
      int ret = (str1[i]-'0')*(str2[j] - '0') + carry;
      carry = ret / 10;
      str += (ret % 10) + '0';
    }
    str+=(carry+'0');
    // 破壊的変更
    reverse(str.begin(),str.end());
    // 先頭の0を消す
    int idx;
    for(int i = 0; i < str.size(); i++){
      idx=i;
      if(str[i]!='0'||str.size()-1==i)
        break;
    }
    str=str.substr(idx);
    for(int i = 0; i < base; i++){
      str+='0';
    }
    ret=add(ret,str);
    base++;
  }
  return ret;
}

int N;
int main(){
  cin>>N;
  string s="1";
  for(int i=1;i<=N+1;i++){
    stringstream ss;
    ss<<i;
    s=mul(s,ss.str());
  }
  s=add(s,"2");
  cout<<s<<endl;
  for(int i=2;i<=N+1;i++)
    cout<<i<<endl;
  return 0;
}