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.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	
	static final int MAX_NUM_OF_PRIME_NUMBERS = 10000;
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		while(true){
			int numOfPrimeNumbers = Integer.parseInt(br.readLine());
			if( numOfPrimeNumbers == 0 )	break;
			System.out.println(getSumOfPrimeNumbers(numOfPrimeNumbers));
		}
	}
	public static long getSumOfPrimeNumbers(int numOfPrimeNumbers){
		long sum = 0;
		long[] primeNumber = getPrimeNumbers();
		for(int i = 0; i < numOfPrimeNumbers; i++){
			sum += primeNumber[i];
		}
		return sum;
	}
	private static long[] getPrimeNumbers(){
		long[] primeNumber = new long[MAX_NUM_OF_PRIME_NUMBERS];
		int numPrimeNumbers = 0;
		long n = 1;
		while(numPrimeNumbers < MAX_NUM_OF_PRIME_NUMBERS){
			if( isPrimeNumber(++n) ){
				primeNumber[numPrimeNumbers] = n;
				numPrimeNumbers++;
			}
		}
		return primeNumber;
	}
	private static boolean isPrimeNumber(long target){
		if( target == 2 )	return true;
		if( target % 2 == 0 )	return false;
		for(int i = 3; i * i < target + 1; i = i + 2){
			if( target % i == 0){
				return false;
			}
		}
		return true;
	}
}