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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = null;

		// エラトステネスの篩
		int MAX = 1000000;
		int num = (MAX - 3) / 2;
		boolean[] prime = new boolean[MAX + 1];
		Arrays.fill(prime, true);
		int count = 1;
		int[] total = new int[10000 + 1];
		total[1] = 2;

		for (int i = 0; i < num; i++) {
			if (prime[i] == true) {
				int primeNum = i + i + 3;
				count++;
				total[count] = total[count - 1] + primeNum;
				for (int j = i + primeNum; j <= MAX; j += primeNum)
					prime[j] = false;
				if(count == 10000) break;
			}
		}
		
		try {
			while ((line = br.readLine()) != null) {
				line = line.trim();
				if (line.equals("0"))
					break;
				int N = Integer.parseInt(line);
				
				System.out.println(total[N]);
			}
			br.close();
		} catch (IOException e) {
			e.printStackTrace();
		}

	}

}