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.*;
public class Main {
	static boolean prime[] = new boolean[200000];
	static int[] result = new int[10002];
	static void Do(int k){
		int sum=0;
		for(int i=1;i<=k;i++){
			sum+=P(i);
		}
	}
	static int P(int n){
		int cnt=0;
		for(int i=0;i<prime.length;i++){
			if(prime[i]){
				cnt++;
				if(cnt==n)return i;
			}
		}
		return 0;
	}
	
	static void doIt(int n){
		if(n==0)return;
		result[n]=result[n-1]+P(n);
	}
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		for(int i=0;i<prime.length;i++)prime[i]=true;
		prime[0]=false;prime[1]=false;
		for(int s=0;s<prime.length;s++){
			if(prime[s])
				for(int i=s+s;i<prime.length;i+=s)prime[i]=false;
		}
		for(int i=1;i<result.length;i++)doIt(i);
		while(in.hasNext()){
			int x = in.nextInt();
			if(x==0)break;
			System.out.println(result[x]);
		}
	}
}