Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.

Given n, the sum of p (i) from i = 1 to n s

s = p (1) + p (2) + .... + p (n)

Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.



Input

Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.

Output

For n in each dataset, print s on one line.

Example

Input

2
9
0


Output

5
100
import java.io.*;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;
public class Main {
	static ArrayList<Integer> list = new ArrayList<Integer>();
	static ArrayList<Integer> listX = new ArrayList<Integer>();
	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		makePrimeList(104730);
		makeSumList();
		while(true) {
			int n = stdIn.nextInt();
			if(n == 0) break;
			
			System.out.println(listX.get(n-1));
		}
	}
	public static void makeSumList() {
		int sum = 0;
		for(int i = 0; i < 10000; i++) {
			sum += list.get(i);
			listX.add(sum);
		}
	}
	public static void makePrimeList(int n) {
		boolean[] isntPrime = new boolean[n];
		isntPrime[0] = isntPrime[1] = true;
		list.add(2);
		for(int i = 3; i < n; i+= 2) {
			if(!isntPrime[i]) {
				list.add(i);
				for(int j = i + i; j < n; j+= i) {
					isntPrime[j] = true;
				}
			}
		}
	}

	
	
}