// Original C++ solution (slower):
#include<bits/stdc++.h>

#include<algorithm>

#include<math.h>

#include<string>

#include<iostream>

#include<vector>

#include<cstdlib>

#include<stdio.h>

#include<cmath>

#include<limits>

#include<iomanip>



using namespace std;



template <class T> using V = vector<T>;



template<class T> inline bool chmin(T& a, T b) {

    if (a > b) {

        a = b;

        return true;

    }

    return false;

}

template<class T> inline bool chmax(T& a, T b) {

    if (a < b) {

        a = b;

        return true;

    }

    return false;

}



const long long INF = 1LL << 60;



using ll = long long;

using db = long double;

using st = string;

using ch = char;

using bl = bool;

using vll = V<ll>;

using vpll =V<pair<ll,ll>>;

using vst = V<st>;

using vdb = V<db>;

using vch = V<ch>;

using vbl = V<bl>;

using graph = V<V<int>>;

using pq = priority_queue<ll>;







#define FOR(i,a,b) for(ll i=(a);i<(b);i++)

#define rFOR(i,a,b) for(ll i=(a);i>=(b);i--)

#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)

#define bgn begin()

#define en end()

#define SORT(a) sort((a).bgn,(a).en)

#define REV(a) reverse((a).bgn,(a).en)

#define M(a,b) max(a,b)

#define rM(a,b) min(a,b)

#define fi first

#define se second

#define sz size()

#define gcd(a,b) __gcd(a,b)

#define co(a) cout<<a<<endl;

#define pb(a) push_back(a);







ll Sum(ll n) {

ll m=0;

  while(n){

    m+=n%10;

    n/=10;

  }

  return m;

}







const int MAX = 510000;

const int MOD = 1000000007;

 

long long fac[MAX], finv[MAX], inv[MAX];

 

void Comuse() {

    fac[0] = fac[1] = 1;

    finv[0] = finv[1] = 1;

    inv[1] = 1;

    for (int i = 2; i < MAX; i++){

        fac[i] = fac[i - 1] * i % MOD;

        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;

        finv[i] = finv[i - 1] * inv[i] % MOD;

    }

}





#define comuse Comuse()



ll combi(int n, int k){

    if (n < k) return 0;

    if (n < 0 || k < 0) return 0;

    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;

}



ll perm(int n,int k){

    if(n < k) return 0;

    if(n < 0 || k < 0) return 0;

    return fac[n] * (finv[k] % MOD) % MOD;

}





ll lcm(ll a,ll b){

  ll n;

  n=a/gcd(a,b)*b;

  return n;

}





vll div(ll n){

  vll ret;

  for(ll i=1;i*i<=n;i++){

    if(n%i==0){

      ret.push_back(i);

      if(i*i!=n){

        ret.push_back(n/i);

      }

    }

  }

  SORT(ret);

  return (ret);

}



ll modpow(ll a,ll n,ll mod){

  ll ans=1;

  while(n>0){

    if(n&1){

      ans=ans*a%mod;

    }



    a=a*a%mod;

    

    n>>=1;

  }



  return ans;

}



void Solve();





const int MAX_N = 131072;

//segment tree 

int NN;

int seg[MAX_N*2-1];

void seguse(){

  for(int i=0;i<2*NN-1;i++){

    seg[i]=INT_MAX;

  }

}



signed main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  cout<<setprecision(20)<<fixed;

  Solve();

}





/****************************************\

| Thank you for viewing my code:)        |

| Written by RedSpica a.k.a. RanseMirage |

| Twitter:@asakaakasaka                  | 

\****************************************/

//segtreeの葉の先頭の添え字はn-1

void Solve(){

  //vll segtree(524287);

  //ll N=262144;

  //5で割って1余るような素数で構成する

  ll n;

    cin>>n;

  ll m=n;

  vll ans(0);

  //ll now=11;

    FOR(i,3,55556){

      bool b=true;

      FOR(j,2,i){

        if(i%j==0){

          b=false;

          break;

        }

      }

      if(b&&i%5==1){

        ans.push_back(i);

        n--;

      }

      if(n==0){

        break;

      }

    }

  



  FOR(i,0,m){

    cout<<ans[i]<<" ";

  }

  cout<<"\n";

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<math.h>

#include<bits/stdc++.h>



using namespace std;



template <class T> using V = vector<T>;



using ll = long long;

using vll = V<ll>;



#define FOR(i,a,b) for(ll i=(a);i<(b);i++)



void Solve();



signed main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  cout<<setprecision(20)<<fixed;

  Solve();

}





/****************************************\

| Thank you for viewing my code:)        |

| Written by RedSpica a.k.a. RanseMirage |

| Twitter:@asakaakasaka                  | 

\****************************************/





void Solve(){

  ll n;

    cin>>n;

  vll ans(0);

  ll count=n;

  FOR(i,2,55555){

    bool b=true;

    FOR(j,2,sqrt(i)+1){

      if(i%j==0){

        b=false;

        break;

      }

    }

    if(b&&i%5==1){

      ans.push_back(i);

      count--;

    }



    if(count==0){

      break;

    }

  }



  FOR(i,0,n){

    cout<<ans[i]<<" ";

  }

  cout<<"\n";

}

// Fastest known solution for this problem: