import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
    static class InputReader {
        BufferedReader reader ;
        StringTokenizer tokenizer ;
        public  InputReader (InputStream in){
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }
        public boolean hasNext(){
            while (tokenizer == null || !tokenizer.hasMoreTokens()){
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                }catch (Exception e) {
                    return false;
                }
             }
            return true;
        }
        public String next(){
            if(hasNext()) return tokenizer.nextToken();
            return null;
        }
        public Integer nextInt() {
            return Integer.parseInt(next());
        }
        public Long nextLong() {
            return Long.parseLong(next());
        }
        public Double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Task task = new Task();
        task.solve(in, out);
        out.flush();
    }
    static class Task {
        int N;
        public long getPsum(int p){
            long sum = 0;
            long tmp = p;//在计算tmp * p 时会出现溢出情况，而导致在和N 比较时出现问题。所以必须使用long
            while (tmp <= N) {
                sum += N / tmp;
                tmp *= p;
            }
            return sum;
        }

        public void solve(InputReader in, PrintWriter out) {
            N = in.nextInt();

            long[] s = new long[N + 1];
            int[] ss = new int[N + 1];
            int[] tmp = new int[N + 1];

            int m = 0;

            //（1）求出1~N的质数
            int[] vis = new int[N + 1];
            for(int i = 2; i <= N; i ++) {
                if(vis[i] == 1) continue;
                //（2）计算每个质数的个数在N！中
                ss[m] = i;
                s[m ++] = getPsum(i);
                for(int j = i; j <= N / i; j ++){
                    vis[j * i] = 1;
                }
            }


            for(int i = 0; i < m; i ++) {
                out.println(ss[i] +" " + s[i]);
            }
        }
    }
}