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.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	static boolean[] isPrime = new boolean[1000000];
	static int[] a;
	static int[] sum;
	static{
		Arrays.fill(isPrime, true);
		isPrime[0]=isPrime[1]=false;
		int count=0;
		int[] b = new int[100000];
		for(int i = 2; i*i<=isPrime.length; i++){
			if(isPrime[i]){
				
				for(int j = i+i; j < isPrime.length; j+=i){
					isPrime[j]=false;
				}
			}
		}
		
		for(int i = 0; i < isPrime.length; i++){
			if(isPrime[i]){
				b[count++]=i;
			}
		}
		a = new int[count];
		sum = new int[count+1];
		for(int i = 0; i < count; i++){
			
			a[i]=b[i];
			sum[i+1] = sum[i]+a[i];
		}
	}
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		while(true){
			int n = cin.nextInt();
			if(n==0){
				break;
			}
			System.out.println(sum[n]);
			
		}
	}

}